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7623fa9 Jan 18, 2017
@jemmybutton @dougmencken @bpiddubnyi
12549 lines (10805 sloc) 501 KB
\switchtobodyfont[11pt]
\definefontfeature
[default]
[default]
[protrusion=quality,expansion=quality]
\setupalign[hz,hanging]
\definepapersize[custom]
[width=153mm, height=210mm]
\setuppapersize[custom][custom]
%\setuppapersize[A5][A5]
\setuppagenumbering[alternative=doublesided]
\setuplayout[backspace=14mm,
width=80mm, height=180mm,
header=5mm,
headerdistance=8mm,
topspace=10mm,
footer=0mm,
margin=52mm]
\definelayout[title][backspace=15mm,
width=123mm, height=180mm,
header=5mm,
headerdistance=8mm,
topspace=10mm,
footer=0mm,
margin=52mm]
\input preamble.tex
\input preamble_be.tex
\usesymbols[cc]
\starttext
\setuplayout[title]
\setupheader [state=stop]
\startalignment[middle]
{\tfb \WORD{The first six books of}}
\vskip 0.25\baselineskip
{\tfc \WORD{The elements of Euclid}}
\vskip 0.5\baselineskip
{\WORD{in which coloured diagrams and symbols are used instead of letters for the greater ease of learners}}
\vskip 0.75\baselineskip
{\tfb \WORD{By Oliver Byrne}}
%{\WORD{SURVEYOR OF HER MAJESTY'S SETTLEMENTS IN THE FALKLAND ISLANDS AND AUTHOR OF NUMEROUS MATHEMATICAL WORKS}}
\defineNewPicture{
scaleFactor := 7/6;
pair A, B, C, D, E, F, G, H, I, J, K, L, M, d[];
A := (0, 0);
B := A shifted (-7/10u, -8/7u);
C = whatever[A, A shifted ((A-B) rotated 90)] = whatever[B, B shifted dir(0)];
d1 := (B-A) rotated -90;
D := A shifted d1;
E := B shifted d1;
d2 := (A-C) rotated -90;
F := C shifted d2;
G := A shifted d2;
d3 := (C-B) rotated -90;
H := B shifted d3;
I := C shifted d3;
J = whatever[A, A shifted dir(90)];
J = whatever[B, C];
K = whatever[A, A shifted dir(90)];
K = whatever[H, I];
L = whatever[B, F];
L = whatever[A, C];
M = whatever[A, I];
M = whatever[B, C];
draw byPolygon(A,B,E,D)(black);
draw byPolygon(L,A,G,F)(byred);
draw byPolygon(C,L,F)(byred);
draw byPolygon(J,M,I,K)(byblue);
draw byPolygon(M,C,I)(byblue);
draw byPolygon(B,J,K,H)(byyellow);
draw byAngle(F, C, A, byyellow, 0);
draw byAngle(B, C, I, byyellow, 0);
draw byAngle(A, C, B, black, 0);
draw byLine(A, K, black, 1, 0);
draw byLineFull(B, F, black, 0, 0)(C, G, 1, 1, 0);
draw byLineFull(A, I, black, 0, 0)(C, H, 1, 1, 0);
draw byLineFull(C, F, byblue, 1, 0)(C, F, 0, 0, 1/2);
draw byLineFull(C, I, byred, 1, 0)(C, I, 0, 0, -1/2);
byLineDefine(A, B, byyellow, 0, 0);
byLineDefine(B, C, byred, 0, 0);
byLineDefine(C, A, byblue, 0, 0);
draw byNamedLineSeq(1/2)(AB,CA,BC);
byLineDefineWithName (C, A, black, 0, 0)(CAb);
byLineStylize (M, M, 1, 0, -1) (CAb);
byLineDefineWithName (A, M, black, 0, 0)(AMb);
byLineStylize (C, C, 0, 1, -1) (AMb);
byLineDefineWithName (B, C, black, 0, 0)(BCb);
byLineStylize (L, L, 0, 1, -1) (BCb);
byLineDefineWithName (L, B, black, 0, 0)(BLb);
byLineStylize (C, C, 1, 0, -1) (BLb);
}
\vfill\vfill
~\hfill\reuseMPgraphic{\currentInstance::currentPicture}\hfill~
\vfill\vfill\vfill
{\tfa github.com/jemmybutton}
\vskip 0.25\baselineskip
{\tfb 2017 ed.\,0.1b}
\vskip \baselineskip
\symbol[cc][cc] \symbol[cc][by] \symbol[cc][sa]
\startnarrower
\setuplocalinterlinespace[line=2ex]
{\tfxx This rendition of Oliver Byrne's \quotation{The first six books of the Elements of Euclid} is made by Slyusarev Sergey and is distributed under CC-BY-SA 4.0 license}
\stopnarrower
\vskip -\baselineskip
\stopalignment
\pagebreak\ \pagebreak
\setupheader[state=start]
\setuplayout[reset]
\startintro[title={Introduction}]
\regularLettrine{T}he arts and sciences have become so extensive, that to facilitate their acquirement is of as much importance as to extend their boundaries. Illustration, if it does not shorten the time of study, will at least make it more agreeable. This Work has a greater aim than mere illustration; we do not introduce colors for the purpose of entertainment, or to amuse \emph{by certain combinations of tint and form}, but to assist the mind in its researches after truth, to increase the facilities of introduction, and to diffuse permanent knowledge. If we wanted authorities to prove the importance and usefulness of geometry, we might quote every philosopher since the day of Plato. Among the Greeks, in ancient, as in the school of Pestalozzi and others in recent times, geometry was adopted ad the best gymnastic of the mind. In fact, Euclid's Elements have become, by common consent, the basis of mathematical science all over the civilized globe. But this will not appear extraordinary, if we consider that this sublime science is not only better calculated than other to call forth the spirit of inquiry, the elevate mind, and to strengthen the reasoning faculties, but also it forms the best introduction to most useful and important vocations of human life. Arithmetic, land-surveying, hydrostatics, pneumatics, optics, physical astronomy, etc. are all dependent on the propositions of geometry.
\kerncharacters[-0.01]{Much however depends on the first communication of any science to a learner, though the best and most easy methods are seldom adopted. Propositions are placed before a student, who though having a sufficient understanding, is told just as much about them on entering at the very threshold of the science, as given him a prepossession most unfavorable to his future study of this delightful subject; or \quotation{the formalities and paraphrenalia of rigour are so ostentatiously put forward, as almost to hide the reality. Endless and perplexing repetitions, which do not confer great exactitude on the reasoning, render the demonstrations involved and obscure, and conceal from the view of the student the consecution of evidence.} Thus an aversion is created in the mind of the pupil, and a subject so calculated to improve the reasoning powers, and give the habit of close thinking, is degraded by a dry and rigid course of instruction into an uninteresting exercise of the memory. To rise the curiosity, and to awaken the listless and dormant powers of younger minds should be the aim of every teacher; but where examples of excellence are wanting, the attempts to attain it are but few, while eminence excites attention and produces imitation. The object of this Work is to introduce a method of teaching geometry, which has been much approved of by many scientific men in this country, as well as in France and America. The plan here adopted forcibly appeals to the eye, the most sensitive and the most comprehensive of our external organs, and its pre-eminence to important subject on the mind is supported by the incontrovertible maxim expressed in well known words of Horace:—}
\vskip 0.5\baselineskip
\startalignment[middle]
\emph{Segnius irritant animos demissa per aurem\\
Quam quae sunt oculis subjecta fidelibus}\\
A feebler impress through the ear is made,\\
Than what is by the faithful eye conveyed.
\stopalignment
\vskip 0.5\baselineskip
\kerncharacters[-0.01]{All language consists of representative signs, and those signs are the best which effect their purposes with the greatest precision and dispatch. Such for all common purposes are the audible signs called words, which are still considered as audible, whether addressed immediately to the ear, or through the medium of letters to the eye. Geometrical diagrams are not signs, but the materials of geometrical science, the object of which is to show the relative quantities of their parts by a process of reasoning called Demonstration. This reasoning has been generally carried on by words, letters, and black or uncoloured diagrams; but as the use of coloured symbols, signs, and diagrams in the linear arts and sciences, renders the process of reasoning more precise, and the attainment more expeditious, they have been in this instance accordingly adopted.}
\kerncharacters[0.01]{Such is the expedition of this enticing mode of communicating knowledge, that the Elements of Euclid can be acquired in less that one third of the time usually employed, and the retention by the memory is much more permanent; these facts have been ascertained by numerous experiments made by the inventor, and several others who have adopted his plans. The particulars of which are few and obvious; the letters annexed to points, lines, or other parts of a diagram are in fact but arbitrary names, and represent them in the demonstration; instead of these, the parts being differently coloured, are made to name themselves, for their forms in corresponding colours represent them in the demonstration.}
\defineNewPicture{
pair A, B, C;
B := (0, 0);
A := B shifted (dir(-145)*3u);
C = whatever[A, A shifted (1,0)] = whatever[B, B shifted dir(-145+90)];
draw byAngleWithName(A, B, C, byyellow, 0)(B);
draw byAngleWithName(B, C, A, byblue, 0)(C);
draw byAngleWithName(C, A, B, byred, 0)(A);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(B, C, byred, 0, 0);
byLineDefine(C, A, byyellow, 0, 0);
draw byNamedLineSeq(0)(AB,BC,CA);
label.top(\sometxt{B}, B);
label.rt(\sometxt{C}, C);
label.lft(\sometxt{A}, A);
}
\drawCurrentPictureInMargin
In order to give a better idea of this system, and of advantages gained by its adoption, let us take a right angled triangle, and express some of its properties both by colours and the method generally employed.
\vskip \baselineskip
\startalignment[middle]
\emph{Some of the properties of the right angled triangle ABC, expressed by the method generally employed:}
\stopalignment
\startitemize[m,joinedup,nowhite]
\item The angle BAC, together with the angles BCA and ABC are equal to two right angles, or twice the angle ABC.
\item The angle CAB added to the angle ACB will be equal to the angle ABC.
\item The angle ABC is greater than either of the angles BAC or BCA.
\item The angle BCA or the angle CAB is less than the angle ABC.
\item If from the angle ABC, there be taken the angle BAC, the remainder will be equal to the angle ACB.
\item The square of AC is equal to the sum of the squares of AB and BC.
\stopitemize
\vskip \baselineskip
\startalignment[middle]
\emph{The same properties expressed by colouring the different parts:}
\stopalignment
\startitemize[m,joinedup,nowhite]
\item $\drawAngle{A} + \drawAngle{B} + \drawAngle{C} = 2 \drawAngle{B} = \drawTwoRightAngles$. \\ That is, the red angle added to the yellow angle added to the blue angle, equal twice the yellow angle, equal two right angles.
\item $\drawAngle{A} + \drawAngle{C} = \drawAngle{B}$. \\ Or in words, the red angle added to the blue angle, equal the yellow angle.
\item $\drawAngle{B} > \drawAngle{A} \mbox{ or } > \drawAngle{C}$. \\ The yellow angle is greater than either the red or blue angle.
\item $\drawAngle{A} \mbox{ or } \drawAngle{C} < \drawAngle{B}$. \\ Either the red or blue angle is less that the yellow angle.
\item $\drawAngle{B} \mbox{ minus } \drawAngle{C} = \drawAngle{A}$. \\ In other terms, the yellow angle made less be the blue angle equal red angle.
\item $\drawUnitLine{CA}^2 = \drawUnitLine{AB}^2 + \drawUnitLine{BC}^2$. \\ That is, the square of the yellow line is equal to the sum of the squares of the blue and red lines.
\stopitemize
\vfill\pagebreak
In oral demonstrations we gain with colours this important advantage, the eye and the ear can be addressed at the same moment, so that for teaching geometry, and other linear arts and sciences, in classes, the system is best ever proposed, this is apparent from the examples given.
\kerncharacters[-0.015]{Whence it is evident that a reference from the test to the diagram is more rapid and sure, by giving the forms and colours of the parts, or by naming the parts and their colours, than naming the parts and letters on the diagram. Besides the superior simplicity, this system is likewise conspicuous for concentration, and wholly excludes the injurious though prevalent practice of allowing the student to commit the demonstration to memory; until reason, and fact, and proof only make impressions of the understanding.}
Again, when lecturing on the principles or properties of figures, if we mention the colour of the part or parts referred to, as in saying, the red angle, the blue line, or lines, etc, the part or parts thus named will be immediately seen by all the class at the same instant; not so if we say the angle ABC, the triangle PFQ, the figure EGKt, and so on; for the letters must be traced one by one before students arrange in their minds the particular magnitude referred to, which often occasions confusion and error, as well as loss of time. Also if the parts which are given as equal, have the same colours in any diagram, the mind will not wander from the object before it; that is, such an arrangement presents an ocular demonstration of the parts to be proved equal, and the learner retains the data throughout the whole of reasoning. But whatever may be the advantages of the present plan, if it be not substituted for, it can always be made a powerful auxiliary to the other methods, for the purpose of introduction, or of a more speedy reminiscence, or of more permanent retention by the memory.
The experience of all who have formed systems to impress facts on the understanding, agree in proving that coloured representations, as pictures, cuts, diagrams, etc. are more easily fixed in the mind than mere sentences unmarked by any peculiarity. Curious as it may appear, poets seem to be aware of this fact more than mathematicians; many modern poets allude to this visible system of communicating knowledge, one of them has thus expressed himself:
\vskip 0.5\baselineskip
\startalignment[middle]
Sounds which address the ear are lost and die\\
In one short hour, but these which strike the eye,\\
Live long upon the mind, the faithful sight\\
Engraves the knowledge with a beam of light.
\stopalignment
\vskip 0.5\baselineskip
This perhaps may be reckoned the only improvement which plain geometry has received since the days of Euclid, and if there were any geometers of note before that time, Euclid's success has quite eclipsed their memory, and even occasioned all good things of that kind to be assigned to him; like Aesop among the writers of Fables. It may also be worthy of remark, as tangible diagrams afford the only medium through which geometry and other linear arts can be taught to the blind, the visible system is no less adapted to the exigencies of the deaf and dumb.
Care must be taken to show that colour has nothing to do with the lines, angles, or magnitudes, except merely to name them. A mathematical line, which is length without breadth, cannot possess colour, yet the junction of two colours on the same plane gives a good idea of what is meant by a mathematical line; recollect we are speaking familiarly, such a junction is to be understood and not the colour, when we say the black line, the red line or lines, etc.
Colours and coloured diagrams may at first appear a clumsy method to convey proper notions of the properties and parts of mathematical figures and magnitudes, however they will be found to afford a means more refined and extensive than any that has hitherto proposed.
We shall here define a point a line, and a surface, and demonstrate a proposition in order to show the truth of this assertion.
A point is that which has position, but not magnitude; or a point is position only, abstracted from the consideration of length, breadth, and thickness. Perhaps the following description is better calculated to explain the nature of mathematical point to those who have not acquired the idea, that the above specious definition.
\defineNewPicture{
angleScale := 2;
pair O, A, B, C;
O := (0, 0);
A := dir(0) scaled 3u;
B := dir(120) scaled 3u;
C := dir(240) scaled 3u;
draw byAngle(A, O, B, byred, 0);
draw byAngle(B, O, C, byblue, 0);
draw byAngle(C, O, A, byyellow, 0);
}
Let three colours \drawCurrentPictureInMargin meet and cover a portion of the paper, where they meet is not blue, nor it is yellow, nor it is red, as it occupies no portion of the plane, for if it did, it would belong to the blue, the red, or the yellow part; yet it exists, and has position without magnitude, so that with a little reflection, this junction of three colours on a plane, gives a good idea of a mathematical point.
A line is length without breadth. With the assistance of colours, nearly in the same manner as before, an idea of a line may be thus given:—
\defineNewPicture{
pair A, B, C, D, E, F;
A := (0, 0);
B := (5/2u, ypart(A));
C := (xpart(A), -2u);
D := (xpart(B), ypart(C));
E := 1/2[A, C];
F := 1/2[B, D];
draw byPolygon(A,B,F,E)(byred);
draw byPolygon(C,D,F,E)(byblue);
}
\drawCurrentPictureInMargin
Let two colours meet and cover a portion of paper; where they meet is not red, nor is it blue; therefore the junction occupies no portion of the plane, and therefore it cannot have breadth, but only length: from which we can readily form an idea of what is meant be a mathematical line. For the purpose of illustration, one colour differing from the colour of the paper, or plane upon which it is drawn, would have been sufficient; hence in future, if we say the red line, the blue line or lines, etc. it is the junctions with the plane upon which they are drawn are to be understood.
\defineNewPicture{
pair A', A'', A''', B', B'', B''', C', C'', C''', D', D'', D''', d[];
d1 := (3/2u, 0);
d2 := (-3/4u, -2/3u);
d3 := (0, -3/2u);
A' := (0, 0);
B' := A' shifted d1;
C' := A' shifted d2;
D' := C' shifted d1;
A'' := A' shifted d3;
B'' := B' shifted d3;
C'' := C' shifted d3;
D'' := D' shifted d3;
A''' := A'' shifted d3;
B''' := B'' shifted d3;
C''' := C'' shifted d3;
D''' := D'' shifted d3;
draw byPolygon(A',B',B'',A'',C'',C')(byred);
draw byPolygon(A'',B'',D'',C'')(byblue);
draw byPolygon(C'',D'',B'',B''',D''',C''')(byyellow);
label.lft(\sometxt{P}, C');
label.lft(\sometxt{R}, C'');
label.rt(\sometxt{S}, B'');
label.rt(\sometxt{Q}, B''');
}
\drawCurrentPictureInMargin
Surface if that which has length and breadth without thickness.
\kerncharacters[-0.02]{When we consider a solid body (PQ), we perceive at once that it has three dimensions, namely :— length, breadth, and thickness; suppose one part of this solid (PS) to be red, and the other part (QR) yellow, and that the colours be distinct without commingling, the blue surface (RS) which separates these parts, or which is the same thing, that which divides the solid without loss of material, must be without thickness, and only possesses length and breadth; this plainly appears from reasoning, similar to that just employed in defining, or rather describing a point and a line.}
The proposition which we have selected to elucidate the manner in which the principles are applier, is the fifth of the first Book.
\defineNewPicture[1/4]{
angleScale := 5/6;
pair A, B, C, D, E;
A := (0, 0);
B := A shifted (u, -2u);
C := B xscaled -1;
D := 9/5[A,B];
E := 9/5[A,C];
draw byAngle(B, A, C, black, 0);
draw byAngle(A, B, C, byblue, 0);
draw byAngle(B, C, A, byblue, 0);
draw byAngle(C, B, E, byyellow, 0);
draw byAngle(D, C, B, byyellow, 0);
draw byAngle(B, D, C, byred, 0);
draw byAngle(C, E, B, byred, 0);
byAngleDefine(E, B, D, black, 1);
byAngleDefine(D, C, E, black, 1);
byLineDefine(B, D, byyellow, 0, 0);
byLineDefine(C, E, byyellow, 0, 0);
byLineDefine(B, E, byblue, 0, 0);
byLineDefine(C, D, byblue, 0, 0);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(A, C, byred, 0, 0);
byLineDefine(B, C, black, 0, 0);
draw byNamedLineSeq(0)(CD,noLine,BC,noLine,BE,CE,AC,AB,BD);
label.top(\sometxt{A}, A);
label.lft(\sometxt{C}, C);
label.rt(\sometxt{B}, B);
label.lft(\sometxt{E}, E);
label.rt(\sometxt{D}, D);
}
\drawCurrentPictureInMargin
In an isosceles triangle ABC. the internal angles at the base ABC, ACB are equal, and when the sides AB, AC are produced, the external angles at the base BCE, CBD are also equal.
\startCenterAlign
Produce \drawUnitLine{AB} and \drawUnitLine{AC},\\
make $\drawUnitLine{BD} = \drawUnitLine{CE}$, draw \drawUnitLine{BE} and \drawUnitLine{CD}.\\
In
\drawFromCurrentPicture{
draw byNamedAngle(BAC);
draw byNamedLineSeq(0)(BE,CE,AC,AB);
}
and
\drawFromCurrentPicture{
draw byNamedAngle(BAC);
draw byNamedLineSeq(0)(BD,CD,AC,AB);
}\\
we have $\drawUnitLine{AB,BD} = \drawUnitLine{AC,CE}$,\\
\drawAngle{BAC} common and $\drawUnitLine{AB} = \drawUnitLine{AC}$:\\
$\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$\\
and $\drawAngle{CEB} = \drawAngle{BDC}$ \inprop[prop:I.IV].\\
Again in \drawLine{BC,BE,CE} and \drawLine{BC,BD,CD},\\
$\drawUnitLine{BD} = \drawUnitLine{CE}$, $\drawAngle{CEB} = \drawAngle{BDC}$\\
and $\drawUnitLine{BE} = \drawUnitLine{CD}$;\\
$\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$
and $\drawAngle{DCB} = \drawAngle{CBE}$ \inprop[prop:I.IV]\\
But $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\therefore \drawAngle{BCA} = \drawAngle{ABC}$.
\stopCenterAlign
\qedNB
\startalignment[middle]
\emph{By annexing Letters to the Diagram.}
\stopalignment
Let the equal sides AB and AC be produced through the extremities BC, of the third side, and in the produced part BD of either, lat any point D be assumed, and from the let AE be cut off equal to AD \inprop[prop:I.III]. Let points E and D, so taken in the produced sides, be connected by straight lines DC and BE with the alternate extremities of the third side of the triangle.
In the triangles DAC and EAB the sides DA and AC are respectively equal to EA and AB, and the included angle A is common to both triangles. Hence \inprop[prop:I.IV] the line DC is equal to BE, the angle ADC to the angle AEB, and the angle ACD to the angle ABE; if from the equal lines AD and AE the equal sides AB and AC be taken, the remainders BD and CE will be equal. Hence in the triangles BDC and CEB, the sides BD and DC are respectively equal to CE and EB, and the angles D and E included by those sides are also equal. Hence \inprop[prop:I.IV] the angles DBC and ECB, which are those included by the third side BC and the productions of the equal sides AB and AC are equal. Also the angles DCB and EBC are equal if those equals be taken from the angles DCA and EBA before proved equal, the remainders, which are the angles ABC and ACD opposite to the equal sides, will be equal.
Therefore in an isosceles triangle, etc.
\qedNB
Our object in this place being to introduce system rather than to teach any particular set of propositions, we have therefore selected the foregoing out of the regular course. For schools and other public places of instruction, dyed chalks will answer to describe the diagrams, etc. for private use coloured pencils will be found very convenient.
We are happy to find that the Elements of Mathematics now forms a considerable part of every sound female education, therefore we call the attention to those interested or engaged in the education of ladies to this very attractive mode of communicating knowledge, and to the succeeding work for its future developement.
We shall for the present conclude by observing, as the senses of sight and hearing can be so forcibly and instantaneously addressed alike with one thousand as with one, the million might be taught geometry and other branches of mathematics with great ease, this would advance the purpose of education more than any thing might be named, for it would teach the people how to think, and not what to think; it is in this particular the great error of education originates.
\vfill\pagebreak
\startsupersection[title={Definitions}]
\startDefinitionOnlyNumber[reference=def:I]
\startalignment[last]
A \emph{point} is that which has no parts.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:II]
\startalignment[last]
A \emph{line} is length without breadth.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:III]
\startalignment[last]
The extremities of a line are points.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:IV]
\startalignment[last]
The straight or right line is that which lies evenly between its extremities.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:V]
\startalignment[last]
A surface is that which has length and breadth only.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:VI]
\startalignment[last]
The extramities of a surface are lines.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:VII]
\startalignment[last]
A plane surface is that which lies evenly between its extremities.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:VIII]
\startalignment[last]
A plane angle is the inclination of two lines to one another, in a plane, which meet together, but are not in the same direction.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C;
A := (0, 0);
B := (u, u);
C := (3/2u, ypart(A));
draw byAngleWithName(B, A, C, byyellow, 0)(A);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(A, C, byred, 0, 0);
draw byNamedLineSeq(0)(AC,AB);
}
\startDefinitionOnlyNumber[reference=def:IX]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A plane rectilinear angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C, D;
A := (0, 0);
B := (4/3u, 0);
C := (0, u);
D := (-4/3u, 0);
draw byAngle(B, A, C, black, 1);
draw byAngle(D, A, C, black, 1);
draw byLine(D, B, black, 0, 0);
draw byLine(A, C, black, 0, 0);
}
\startDefinitionOnlyNumber[reference=def:X]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
When one straight line standing on another straight line makes the adjacent angles equal, each of these angles is called a \emph{right angle}, and each of these lines is said to be \emph{perpendicular} to one another
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C;
A := (0, 0);
B := (-u, u);
C := (3/2u, ypart(A));
draw byAngleWithName(B, A, C, byred, 0)(A);
byLineDefine(A, B, byyellow, 0, 0);
byLineDefine(A, C, byblue, 0, 0);
draw byNamedLineSeq(0)(AC,AB);
}
\drawCurrentPictureInMargin[inside]
\startDefinitionOnlyNumber[reference=def:XI]
\startalignment[last]
An obtuse angle is an angle greater than right angle
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C;
A := (0, 0);
B := (u, u);
C := (3/2u, ypart(A));
draw byAngleWithName(B, A, C, byblue, 0)(A);
byLineDefine(A, B, byyellow, 0, 0);
byLineDefine(A, C, byred, 0, 0);
draw byNamedLineSeq(0)(AC,AB);
}
\drawCurrentPictureInMargin[inside]
\startDefinitionOnlyNumber[reference=def:XII]
\startalignment[last]
An acute angle is an angle less than a right angle.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:XII]
\startalignment[last]
A term or boundary is the extremity of any thing.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:XII]
\startalignment[last]
A figure is a surface enclosed on all sides by a line or lines.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair O, A, B, C, D, E;
numeric r;
r := 2/3u;
O := (0, 0);
A := dir(0) scaled r;
B := dir(60) scaled r;
C := dir(130) scaled r;
D := dir(180) scaled r;
E := dir(-60) scaled r;
draw byLine(O, B)(black, 0, 0);
draw byLine(O, C)(byred, 0, 0);
draw byLine(O, E)(byyellow, 0, 0);
draw byLine(D, A)(byblue, 0, 0);
draw byCircle(O, r, byred, 0, 0, 0)(O);
}
\startDefinitionOnlyNumber[reference=def:XV]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A circle is a plane figure, bounded by one continued line, called its circumference or periphery; and having a certain point within it, from which all straight lines drawn to its circumference are equal.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:XVI]
\startalignment[last]
This point (from which the equal lines are drawn) is called the centre of the circle.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair O, A, B;
numeric r;
r := 3/4u;
O := (0, 0);
A := dir(0) scaled r;
B := dir(180) scaled r;
draw byLine(A, B)(byyellow, 0, 0);
draw byCircle(O, r, byred, 0, 0, 0)(O);
}
\startDefinitionOnlyNumber[reference=def:XVII]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A diameter of a circle is a straight line drawn through the centre, terminated both ways in the circumference.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair O, A, B;
numeric r;
r := 3/4u;
O := (0, 0);
A := dir(0) scaled r;
B := dir(180) scaled r;
draw byLine(A, B)(byblue, 0, 0);
draw byArc(O, r, 0, 4, byyellow, 0, 0, 0, 0)(O);
draw byArc(O, r, 4, 8, byyellow, 1, 0, 0, 0)(O);
}
\startDefinitionOnlyNumber[reference=def:XVIII]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A semicircle is the figure contained by the diameter, and the part of the circle cut off by the diameter.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair O, A, B;
path P;
numeric r;
r := 3/4u;
P := fullcircle scaled 2r;
O := (0, 0);
A := point 1 of P;
B := point 3 of P;
draw byLine(A, B)(byred, 0, 0);
draw byArc(O, r, 1, 3, byblue, 0, 0, 0, 0)(O);
draw byArc(O, r, 3, 9, byblue, 1, 0, 0, 0)(O);
}
\startDefinitionOnlyNumber[reference=def:XIX]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A segment of a circle is a figure contained by straight line and the part of the circumference which it cuts off.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:XX]
\startalignment[last]
A figure contained by straight lines only, is called a rectilinear figure.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:XXI]
\startalignment[last]
A triangle is a rectilinear figure included by three sides.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C, D;
A := (0, 0);
B := (u, 1/2u);
C := (-1/2u, -4/3u);
D := (4/3u, -u);
draw byLine(C, B)(byred, 0, 0);
draw byLine(A, D)(byblue, 0, 0);
byLineDefine(A, B, byyellow, 0, 0);
byLineDefine(A, C, byyellow, 0, 0);
byLineDefine(B, D, byyellow, 0, 0);
byLineDefine(C, D, black, 0, 0);
draw byNamedLineSeq(0)(BD,CD,AC,AB);
}
\startDefinitionOnlyNumber[reference=def:XXII]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A quadrilateral figure is one which is bounded by four sides. The straight lines \drawUnitLine{AD} and \drawUnitLine{CB} connecting the vertices of the opposite angles of a quadrilateral figure, are called its diagonals.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:XXIII]
\startalignment[last]
A polygon is a rectilinear figure bounded by more that four sides.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C;
A := dir(-30) scaled 1/2u;
B := dir(-150) scaled 1/2u;
C := dir(90) scaled 1/2u;
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(B, C, byred, 0, 0);
byLineDefine(C, A, byyellow, 0, 0);
draw byNamedLineSeq(0)(AB,BC,CA);
}
\startDefinitionOnlyNumber[reference=def:XXIV]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A triangle whose sides are equal, is said to be equilateral.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C;
A := dir(-60) scaled 1/2u;
B := dir(-120) scaled 1/2u;
C := dir(90) scaled 1/2u;
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(B, C, byred, 0, 0);
byLineDefine(C, A, byred, 0, 0);
draw byNamedLineSeq(0)(AB,BC,CA);
}
\startDefinitionOnlyNumber[reference=def:XXV]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A triangle which has only two sides equal is called an isosceles triangles.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:XXVI]
\startalignment[last]
A scalene triangle is one which has no two sides equal.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C;
A := (0, 0);
B := (-u, 0);
C := (0, 3/4u);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(B, C, byyellow, 0, 0);
byLineDefine(C, A, byblue, 0, 0);
draw byNamedLineSeq(0)(AB,BC,CA);
}
\startDefinitionOnlyNumber[reference=def:XXVII]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A right angled triangle is that which has a right angle.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C;
A := (-1/4u, 0);
B := (-u, 0);
C := (0, 3/4u);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(B, C, byblue, 0, 0);
byLineDefine(C, A, byyellow, 0, 0);
draw byNamedLineSeq(0)(AB,BC,CA);
}
\startDefinitionOnlyNumber[reference=def:XXVIII]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
An obtuse angled triangle is that which has an obtuse angle.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C;
A := (0, 0);
B := (-u, 0);
C := (-1/4u, 3/4u);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(B, C, byyellow, 0, 0);
byLineDefine(C, A, byred, 0, 0);
draw byNamedLineSeq(0)(AB,BC,CA);
}
\startDefinitionOnlyNumber[reference=def:XXIX]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
An acute angled triangle is that which has three acute angles.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C, D;
numeric s;
s := u;
A := (0, 0);
B := (s, 0);
C := (0, s);
D := (s, s);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(A, C, byblue, 0, 0);
byLineDefine(B, D, byyellow, 0, 0);
byLineDefine(C, D, black, 0, 0);
draw byNamedLineSeq(0)(AB,AC,CD,BD);
}
\startDefinitionOnlyNumber[reference=def:XXX]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
Of four-sided figures, a square is that which has all its sides equal, and all its angles right angles.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C, D;
numeric s;
s := u;
A := (0, 0);
B := (s, 0);
C := A shifted (dir(80) scaled s);
D := B shifted (dir(80) scaled s);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(A, C, byblue, 0, 0);
byLineDefine(B, D, byyellow, 0, 0);
byLineDefine(C, D, black, 0, 0);
draw byNamedLineSeq(0)(AB,AC,CD,BD);
}
\startDefinitionOnlyNumber[reference=def:XXXI]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A rhombus is that which has all its sides equal, but its angles are not right angles.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C, D;
numeric s;
s := u;
A := (0, 0);
B := (4/3s, 0);
C := (0, 3/4s);
D := (4/3s, 3/4s);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(A, C, byred, 0, 0);
byLineDefine(B, D, byred, 0, 0);
byLineDefine(C, D, byblue, 0, 0);
draw byNamedLineSeq(0)(AB,AC,CD,BD);
}
\startDefinitionOnlyNumber[reference=def:XXXII]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
An oblong is that which has all its angles right angles, but has not all its sides equal.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C, D;
numeric s;
s := u;
A := (0, 0);
B := (s, 0);
C := (1/4s, 3/4s);
D := (s + 1/4s, 3/4s);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(A, C, byred, 0, 0);
byLineDefine(B, D, byred, 0, 0);
byLineDefine(C, D, byblue, 0, 0);
draw byNamedLineSeq(0)(AB,AC,CD,BD);
}
\startDefinitionOnlyNumber[reference=def:XXXIII]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
A rhomboid is that which has its opposite sides equal to one another, but all its sides are not equal nor its angles right angles.
\stopalignment
\stopDefinitionOnlyNumber
\startDefinitionOnlyNumber[reference=def:XXXIV]
\startalignment[last]
All other quadrilateral figures are called trapezimus.
\stopalignment
\stopDefinitionOnlyNumber
\defineNewPicture{
pair A, B, C, D;
numeric s;
s := u;
A := (0, 0);
B := (4/3s, 0);
C := (0, 1/2s);
D := (4/3s, 1/2s);
draw byLine(A, B, byred, 0, 0);
draw byLine(C, D, byyellow, 0, 0);
}
\startDefinitionOnlyNumber[reference=def:XXXV]
\drawCurrentPictureInMargin[inside]
\startalignment[last]
Parallel straight lines are such as are in the same plane, and which being produced continually in both directions would never meet.
\stopalignment
\stopDefinitionOnlyNumber
\stopsupersection
\startsupersection[title={Postulates}]
\startPostulateOnlyNumber[reference=post:I]
Let it be granted that a straight line may be drawn from any one point to any other point.
\stopPostulateOnlyNumber
\startPostulateOnlyNumber[reference=post:II]
Let it be granted that a finite straight line may be produced to any length in a straight line.
\stopPostulateOnlyNumber
\startPostulateOnlyNumber[reference=post:III]
Let it be granted that a circle may be described with any centre at any distance from that centre.
\stopPostulateOnlyNumber
\stopsupersection
\vfill\pagebreak
\startsupersection[title={Axioms}]
\startAxiomOnlyNumber[reference=ax:I]
Magnitudes which are equal to the same are equal to each other.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:II]
If equals be added to equals the sums will be equal.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:III]
If equals be taken away from equals the remainders will be equal.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:IV]
If equals be added to unequals the sums will be unequal.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:V]
If equals be taken away from unequals the remainders will be unequal.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:VI]
The doubles of the same or equal magnitudes are equal.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:VII]
The halves of the same or equal magnitudes are equal.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:VIII]
The magnitudes which coincide with one another, or exactly fill the same space, are equal.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:IX]
The whole is greater than its part.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:X]
Two straight lines cannot include a space.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:XI]
All right angles are equal.
\stopAxiomOnlyNumber
\startAxiomOnlyNumber[reference=ax:XII]
\defineNewPicture{
pair A, B, C, D, E, F, G, H;
numeric s;
s := 3/2u;
A := (0, 0);
B := (4/3s, 0);
C := (0, s);
D := (4/3s, s);
E := (1/3s, 7/6s);
F := (xpart(E), -1/6s);
G = whatever[A, B] = whatever[E, F];
H = whatever[C, D] = whatever[E, F];
draw byAngleWithName(B, G, E, byred, 0)(G);
draw byAngleWithName(D, H, F, byyellow, 0)(H);
draw byLine(A, B, byblue, 0, 0);
draw byLine(C, D, byred, 0, 0);
draw byLine(E, F, black, 0, 0);
}
\drawCurrentPictureInMargin[inside]
If two straight lines $\left(\vcenter{\nointerlineskip\hbox{\drawUnitLine{AB}}\nointerlineskip\hbox{\drawUnitLine{CD}}}\right)$ meet a third straight line (\drawUnitLine{EF}) so as to make the two interior angles (\drawAngle{H} and \drawAngle{G}) on the same side less than two straight angles, these two straight lines will meet if they be produced on that side on which the angles are less than two right angles.
\stopAxiomOnlyNumber
\stopsupersection
\vfill\pagebreak
\startsupersection[title={Euclidations}]
The twelfth axiom may be expressed in any of the following ways:
\startitemize[m,joinedup,nowhite]
\item Two diverging straight lines cannot be both parallel to the same straight line.
\item If a straight line intersect one of the two parallel straight lines it must also intersect the other.
\item Only one straight line can be drawn through a given point, parallel to a given straight line.
\stopitemize
Geometry has for its principal objects the exposition and explanation of the properties of \emph{figure}, and figure is defined to be the relation which subsists between the boundaries of space. Space or magnitude is of three kinds, \emph{linear}, \emph{superficial}, and \emph{solid}.
\defineNewPicture{
pair A, B, C;
numeric s;
s := 3/2u;
A := (0, s);
B := (1/2s, 0);
C := B xscaled -1;
draw byAngleWithName(B, A, C, byyellow, 0)(A);
byLineDefine(B, A, byblue, 0, 0);
byLineDefine(C, A, byred, 0, 0);
draw byNamedLineSeq(0)(CA,BA);
label.urt(\sometxt{A}, A);
}\drawCurrentPictureInMargin
Angles might properly be considered as a fourth species of magnitude. Angular magnitude evidently consists of parts, and must therefore be admitted to be a species of quantity. The student must not suppose that the magnitude of an angle is affected by the length of the straight lines which include it and of whose mutual divergence it is the measure. The \emph{vertex} of an angle is the point the \emph{sides} of the \emph{legs} of the angle meet, as A.
\defineNewPicture{
pair B, C, D, E, F, G, H;
numeric s;
s := 5/4u;
C := (0, 0);
B := dir(0)*s;
D := dir(50)*s;
E := dir(-30)*s;
F := E scaled -1;
G := D scaled -1;
H := B scaled -1;
angleScale := 4/3;
draw byAngle(E, C, B, byyellow, 0);
draw byAngle(B, C, D, black, 0);
draw byAngle(D, C, F, byblue, 0);
draw byAngle(F, C, H, byred, 0);
draw byAngle(H, C, G, byyellow, 1);
draw byAngle(G, C, E, byblue, 1);
draw byLine(B, H, byblue, 0, 0);
draw byLine(D, G, byred, 0, 0);
draw byLine(E, F, black, 0, 0);
label.bot(\sometxt{C}, C shifted (0, -3pt));
label.bot(\sometxt{B}, B);
label.lrt(\sometxt{D}, D);
label.llft(\sometxt{F}, F);
label.bot(\sometxt{H}, H);
label.lrt(\sometxt{G}, G);
label.llft(\sometxt{E}, E);
}
\drawCurrentPictureInMargin
An angle is often designated by a single letter when its legs are the only lines which meet together at its vertex. Thus the red and blue lines form the yellow angle, which in other systems would be called angle A. But when more than two lines meet in the same point, it was necessary by former methods, in order to avoid confusion, to employ three letters to designate an angle about that point, the letter which marked the vertex of the angle being always placed in the middle. Thus the black and red lines meeting together at C, form the blue angle, and has been usually denominated the angle FCD or DCF. The lines FC and CD are the legs of the angle; the point C is its vertex. In like manner the black angle would be designated the angle DCB or BCD. The red and blue angles added together, or the angle HCD added to FCD, make the angle HCD; and so of other angles.
When the legs of an angle are produced or prolonged beyond its vertex, the angles made by them on both sides of the vertex are said to be \emph{vertically opposite} to each other: thus the red and yellow angles are said to be vertically opposite angles.
\emph{Superposition} is the process by which one magnitude may be conceived to be placed upon another, so as exactly to cover it, or so that every part of each shall exactly coincide.
A line is said to be \emph{produced}, when it is extended, prolonged, or it has length increased, and the increase of length which it receives is called \emph{produced part}, or its \emph{production}.
The entire length of the line or lines which enclose a figure, is called its \emph{perimeter}. The first six books of Euclid treat of plain figures only. A line drawn from the centre of a circle to its circumference, is called a \emph{radius}. That side of a right angled triangle, which is opposite to the right angle, is called the \emph{hypotenuse}. An oblong is defined in the second book, and called a \emph{rectangle}. All lines which considered in the first six books of the Elements are supposed to be in the same plane.
The \emph{straight-edge} and \emph{compasses} are the only instruments, the use of which is permitted in Euclid, or plain Geometry. To declare this restriction is the object of the postulated.
The \emph{Axioms} of geometry are certain general propositions, the truth of which is taken to be self-evident and incapable of being established by demonstration.
\emph{Propositions} are those results which are obtained in geometry by a process of reasoning. There are two species of propositions in geometry, \emph{problems} and \emph{theorems}.
A \emph{Problem} is a proposition in which something is proposed to be done; as a line to be drawn under some given conditions, a circle to be described, some figure to be constructed, etc.
The \emph{solution} of the problem consists in showing how the thing required may be done by the aid of the rule of straight-edge and compasses.
The \emph{demonstration} consists in proving that the process indicated in the solution attains the required end.
A \emph{Theorem} is a proposition in which the truth of some principle is asserted. This principle must be deduced from the axioms and definitions, or other truths previously and independently established. To show this is the object of demonstration.
A \emph{Problem} is analogous to a postulate.
A \emph{Theorem} resembles an axiom.
A \emph{Postulate} is a problem, the solution to which is assumed.
An \emph{Axiom} is a theorem, the truth of which is granted without demonstration.
A \emph{Corollary} is an inference deduced immediately from a proposition.
A \emph{Scholium} is a note or observation on a proposition not containing an inference of sufficient importance to entitle it to the name of \emph{corollary}.
A \emph{Lemma} is a proposition merely introduced for the purpose of establishing some more important proposition.
\stopsupersection
\vfill\pagebreak
\startsupersection[title={Symbols and abbreviations}]
\symb{$\therefore$}
expresses the word \emph{therefore}.
\symb{$\because$}
expresses the word \emph{because}.
\symb{$=$}
expresses the word \emph{equal}. This sign of equality may be read \emph{equal to}, or \emph{is equal to}, or \emph{are equal to}; but the discrepancy in regard to the introduction of the auxiliary verbs \emph{is}, \emph{are}, etc. cannot affect the geometrical rigour.
\symb{$\neq$}
means the same as if the words \emph{`not equal'} were written.
\symb{$>$}
signifies \emph{greater than}.
\symb{$<$}
signifies \emph{less than}.
\symb{$\ngtr$}
signifies \emph{not greater than}.
\symb{$\nless$}
signifies \emph{not less than}.
\symb{$+$}
is read \emph{plus} (\emph{more}), the sign of addition; when interposed between two or more magnitudes, signifies their sum.
\symb{$-$}
is read \emph{minus} (\emph{less}), signifies subtraction; and when placed between two quantities denotes that the latter is taken from the former.
\symb{$\times$}
this sign expresses the product of two or more numbers when placed between them in arithmetic and algebra; but in geometry it is generally used to express a \emph{rectangle}, when placed between \quotation{two straight lines which contain one if its right angles.} A \emph{rectangle} may also be represented by placing a point between two of its conterminous sides.
\symb{$:\ ::\ :$}
expresses an \emph{analogy} or \emph{proportion}; thus is A, B, C and D represent four magnitudes, and A has to B the same ratio that C has to D, the proportion is thus briefly written
$A : B :: C : D$, $A : B = C : D$, or $\dfrac{A}{B} = \dfrac{C}{D}$.
This equality or sameness of ratio is read,
as A is to B, so is C to D;
or A is to B, as C is to D.
\symb{$\parallel$}
signifies \emph{parallel to}.
\symb{$\perp$}
signifies \emph{perpendicular to}.
\defineNewPicture{
pair A, B, C, D;
numeric s;
s := 3/2u;
A := (0, 0);
B := dir(0)*s;
C := dir(50)*s;
D := dir(90)*s;
draw byAngle(B, A, C, white, 0);
draw byAngle(B, A, D, white, 0);
}
\symb{\drawAngle{BAC}}
signifies \emph{angle}.
\symb{\drawAngle{BAD}}
signifies \emph{right angle}.
\symb{\drawTwoRightAngles}
signifies \emph{two right angles}.
\defineNewPicture{
pair A, B, C, D;
A := (0, -1/4u);
B := (u, 0);
C := (-u, 0);
D := (0, u);
byLineDefine (A, D, black, 0, 0);
byLineStylize (A, B, 0, 0, 0) (AD);
byLineDefine (B, D, black, 0, 0);
byLineStylize (B, C, 0, 0, 0) (BD);
byLineDefine (C, D, black, 0, 0);
byLineStylize (C, B, 0, 0, 0) (CD);
}
\symb{\drawFromCurrentPicture{
draw byNamedLine(AD);
draw byNamedLine(BD);
draw byNamedLine(CD);
}
or
\drawFromCurrentPicture{
draw byNamedLine(AD);
draw byNamedLine(BD);
}}
briefly designates a \emph{point}.
A square is described on a line is concisely written thus, $\drawUnitLine{AD}^2$.
In the same manner twice the square of, is expressed by $2 \cdot \drawUnitLine{AD}^2$.
\symb{def.}
signifies \emph{definition}.
\symb{pos.}
signifies \emph{postulate}.
\symb{ax.}
signifies \emph{axiom}.
\symb{hyp.}
signifies \emph{hypothesis}. It may be necessary here to remark, that \emph{hypothesis} is the condition assumed or taken for granted. Thus, the hypothesis of the proposition given in the Introduction, is that the triangle is isosceles, or that its legs are equal.
\symb{const.}
signifies \emph{construction}. The \emph{construction} is the change made in the original figure, by drawing lines, making angles, describing circles, etc. in order to adapt it to the argument of the demonstration or the solution of the problem. The conditions under which these changes are made, are as indisputable as those contained in the hypothesis. For instance, if we make an angle equal to a given angle, these two angles are equal by construction.
\symb{Q. E. D.}
signifies \emph{Quod erat demonstrandum}. Which was to be demonstrated.
\stopsupersection
\stopintro
\startbook[title={Book I}]
\startVerboseProposition[title={Prop. I. Prob.}, reference=prop:I.I]
\defineNewPicture[1/2]{
pair A, B, C;
path P[];
numeric r;
r := 3/2u;
A := (0, 0);
B := (r, 0);
P1 := fullcircle scaled 2r;
P2 := fullcircle scaled 2r shifted B;
C := P1 intersectionpoint P2;
byLineDefine(A, B, black, 0, 0);
byLineDefine(B, C, byred, 0, 0);
byLineDefine(C, A, byyellow, 0, 0);
draw byNamedLineSeq(1)(AB,CA,BC);
draw byCircle(A, r, byblue, 0, 0, 1/2)(A);
draw byCircle(B, r, byred, 0, 0, 1/2)(B);
}
\drawCurrentPictureInMargin
\problemNP{O}{n}{a given finite straight line (\drawUnitLine{AB}) to describe an equilateral triangle.}
\startCenterAlign
Describe \offsetPicture{15pt}{0pt}{\drawFromCurrentPicture{draw byNamedLine (AB); draw byNamedCircle (A);}} and \offsetPicture{15pt}{0pt}{\drawFromCurrentPicture{draw byNamedLine (AB); draw byNamedCircle (B);}}
\inpost[post:III];\\
draw \drawUnitLine{CA} and \drawUnitLine{BC} \inpost[post:I].\\
Then will \drawLine[bottom][triangleABC]{AB,CA,BC} be equilateral.
For $\drawUnitLine{AB} = \drawUnitLine{CA}$ \indef[def:XV];\\
and $\drawUnitLine{AB} = \drawUnitLine{BC}$ \indef[def:XV];\\
$\therefore \drawUnitLine{CA} = \drawUnitLine{BC}$ \inax[post:I];\\
and therefore \triangleABC\ is the equilateral triangle required.
\stopCenterAlign
\qed
\stopVerboseProposition
\startProposition[title={Prop. II. Prob.}, reference=prop:I.II]
\defineNewPicture{
pair A, B, C, D, E, F;
path P[];
numeric r[];
A := (0, 0);
B := (-3/5u, -3/5u);
C := (-2u, -1/3u);
r1 := abs(A-B);
D := (fullcircle scaled 2r1 shifted A) intersectionpoint (fullcircle scaled 2r1 shifted B);
r2 := abs(B-C);
r3 := r1 + r2;
P1 := fullcircle scaled 2r2 shifted B;
P2 := fullcircle scaled 2r3 shifted D;
E := (D -- 10[D, B]) intersectionpoint P1;
F := (D -- 10[D, A]) intersectionpoint P2;
byLineDefine(A, B, black, 1, 0);
byLineDefine(B, C, black, 0, 0);
byLineDefine(B, D, byred, 0, 0);
byLineDefine(D, A, byred, 0, 0);
byLineDefine(B, E, byyellow, 0, 0);
byLineDefine(A, F, byblue, 0, 0);
draw byNamedLineSeq(0)(AF,DA,BD,BE);
draw byNamedLineSeq(0)(AB,BC);
draw byCircle(D, r3, byred, 0, 0, 1/2)(A);
draw byCircle(B, r2, byblue, 0, 0, -1/2)(B);
}
\drawCurrentPictureInMargin
\problemNP{F}{rom}{a given point (\drawUnitLine{DA,AF}) to draw a straight line equal to a given straight line (\drawUnitLine{BC}).}
\startCenterAlign
Draw \drawUnitLine{AB} \inpost[post:I], describe \drawFromCurrentPicture[bottom]{
startTempScale(scaleFactor*3);
draw byNamedLineSeq(0)(AB,BD,DA) rotated (180-lineAngle.AB);
stopTempScale;
} \inprop[prop:I.I],\\
produce \drawUnitLine{DA} \inpost[post:II],\\
describe
\drawFromCurrentPicture{draw byNamedLine (BC); draw byNamedCircle(B);}
\inpost[post:III], and
\drawFromCurrentPicture{draw byNamedLine (BD); draw byNamedLine (BE); draw byNamedCircle(A);}
\inpost[post:III];\\
produce \drawUnitLine{BD} \inpost[post:II],\\
then \drawUnitLine{AF} is the line required.
For $\drawUnitLine{BE,BD} = \drawUnitLine{DA,AF}$ \indef[def:XV],\\
and $\drawUnitLine{BD} = \drawUnitLine{DA}$ (const.),\\
$\therefore \drawUnitLine{BE} = \drawUnitLine{AF}$ \inax[post:III],\\
but \indef[def:XV] $\drawUnitLine{BC} = \drawUnitLine{BE} = \drawUnitLine{AF}$;
$\therefore \drawUnitLine{AF}$ drawn from the given point (\drawUnitLine{DA,AF}), is equal to the given line \drawUnitLine{BC}.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. III. Prob.}, reference=prop:I.III]
\defineNewPicture{
pair A, B, C, D, E, F;
path P;
numeric r;
A := (0, 0);
r := 7/4u;
B := A shifted (r, 0);
C := A shifted (4/3r, 0);
D := A shifted dir(30)*r;
E := A shifted (7/6r, -1/6r);
F := A shifted (7/6r, -7/6r);
byLineDefine(A, B, black, 0, 0);
byLineDefine(B, C, black, 1, 0);
byLineDefine(A, D, byred, 0, 0);
draw byNamedLineSeq(0)(BC,AB,AD);
draw byLine(E, F, byblue, 0, 0);
draw byCircle(A, r, byblue, 0, 0, 0)(A);
}
\drawCurrentPictureInMargin
\problemNP{F}{rom}{the greater (\drawUnitLine{AB,BC}) of two given straight lines, to cut off a part equal to the less (\drawUnitLine{EF}).}
\startCenterAlign
Draw $\drawUnitLine{AD} = \drawUnitLine{EF}$ \inprop[prop:I.II];\\
describe \drawFromCurrentPicture{draw byNamedLine (AD); draw byNamedCircle(A);} \inpost[post:III],\\
then $\drawUnitLine{EF} = \drawUnitLine{AB}$
For $\drawUnitLine{AD} = \drawUnitLine{AB}$ \indef[def:XV],\\
and $\drawUnitLine{EF} = \drawUnitLine{AD}$ (const.);
$\therefore \drawUnitLine{EF} = \drawUnitLine{AB}$ \inax[ax:I];
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. IV. Theor.}, reference=prop:I.IV]
\defineNewPicture{
pair A, B, C, D, E, F, d;
A := (0, 0);
B := A shifted (-5/2u, -7/2u);
C := A shifted (1/3u, -5/2u);
d := (0, -4u);
D := A shifted d;
E := B shifted d;
F := C shifted d;
draw byAngleWithName(B, A, C, byyellow, 0)(A);
draw byAngleWithName(A, B, C, byblue, 0)(B);
draw byAngleWithName(B, C, A, byred, 0)(C);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(B, C, black, 0, 0);
byLineDefine(C, A, byblue, 0, 0);
draw byNamedLineSeq(0)(CA,BC,AB);
draw byAngleWithName(E, D, F, byyellow, 0)(D);
draw byAngleWithName(D, E, F, byblue, 0)(E);
draw byAngleWithName(E, F, D, byred, 0)(F);
byLineDefine(D, E, byred, 0, 1);
byLineDefine(E, F, black, 0, 1);
byLineDefine(F, D, byblue, 0, 1);
draw byNamedLineSeq(0)(FD,EF,DE);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{two triangles have two sides of the one respectively equal to two sides of the other, (\drawUnitLine{AB} to \drawUnitLine{DE} and \drawUnitLine{CA} to \drawUnitLine{FD}) and the angles (\drawAngle{A} and \drawAngle{D}) contained by those equal sides also equal; then their bases or their sides (\drawUnitLine{BC} and \drawUnitLine{EF}) are also equal: and the remaining angles opposite to equal sides are respectively equal ($\drawAngle{B} = \drawAngle{E}$ and $\drawAngle{C} = \drawAngle{F}$): and the triangles are equal in every respect.}
Let two triangles be conceived, to be so placed, that the vertex of one of the equal angles, \drawAngle{A} or \drawAngle{D}; shall fall upon that of the other, and \drawUnitLine{AB} to coincide with \drawUnitLine{DE}, then will \drawUnitLine{CA} coincide with \drawUnitLine{FD} if applied: consequently \drawUnitLine{BC} will coincide with \drawUnitLine{EF}, or two straight lines will enclose a place, which is impossible (ax. 10), therefore $\drawUnitLine{BC} = \drawUnitLine{EF}$, $\drawAngle{B} = \drawAngle{E}$ and $\drawAngle{C} = \drawAngle{F}$, and the triangles \drawLine{CA,BC,AB} and \drawLine{FD,EF,DE} coincide, when applied, they are equal in every respect.
\qed
\stopProposition
\startProposition[title={Prop. V. Theor.}, reference=prop:I.V]
\defineNewPicture{
pair A, B, C, D, E;
picture q;
A := (0, 0);
B := A shifted (u, -2u);
C := B xscaled -1;
D := 9/5[A,B];
E := 9/5[A,C];
draw byAngle(B, A, C, black, 0);
draw byAngle(A, B, C, byblue, 0);
draw byAngle(B, C, A, byblue, 0);
draw byAngle(C, B, E, byyellow, 0);
draw byAngle(D, C, B, byyellow, 0);
draw byAngle(B, D, C, byred, 0);
draw byAngle(C, E, B, byred, 0);
byAngleDefine(E, B, D, black, 1);
byAngleDefine(D, C, E, black, 1);
byLineDefine(B, D, byyellow, 0, 0);
byLineDefine(C, E, byyellow, 0, 0);
byLineDefine(B, E, byblue, 0, 0);
byLineDefine(C, D, byblue, 0, 0);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(A, C, byred, 0, 0);
byLineDefine(B, C, black, 0, 0);
draw byNamedLineSeq(0)(CD,noLine,BC,noLine,BE,CE,AC,AB,BD);
}
\drawCurrentPictureInMargin
\problemNP{I}{n}{any isosceles triangle \drawLine[bottom]{BC,AC,AB} if the equal sides be produced, the external angles at the base are equal, and the internal angles at the base are also equal.}
\startCenterAlign
Produce \drawUnitLine{AB} and \drawUnitLine{AC} \inpost[post:II],\\
take $\drawUnitLine{BD} = \drawUnitLine{CE}$ \inprop[prop:I.III];\\
draw \drawUnitLine{BE} and \drawUnitLine{CD}.
Then in
\drawFromCurrentPicture{
draw byNamedAngle(BAC);
draw byNamedLineSeq(0)(BE,CE,AC,AB);
}
and
\drawFromCurrentPicture{
draw byNamedAngle(BAC);
draw byNamedLineSeq(0)(BD,CD,AC,AB);
}\\
we have $\drawUnitLine{AB,BD} = \drawUnitLine{AC,CE}$ (const.),\\
\drawAngle{BAC} common to both,\\
and $\drawUnitLine{AB} = \drawUnitLine{AC}$ (hyp.)\\
$\therefore \drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\drawUnitLine{BE} = \drawUnitLine{CD}$ and $\drawAngle{CEB} = \drawAngle{BDC}$ \inprop[prop:I.IV].
Again in \drawLine{BE,CE,BC} and \drawLine{BD,CD,BC}\\
we have $\drawUnitLine{BD} = \drawUnitLine{CE}$, $\drawAngle{CEB} = \drawAngle{BDC}$ and $\drawUnitLine{BE} = \drawUnitLine{CD}$,\\
$\therefore \drawAngle{DCE,DCB} = \drawAngle{EBD,CBE}$ and $\drawAngle{DCB} = \drawAngle{CBE}$ \inprop[prop:I.IV]\\
but $\drawAngle{BCA,DCB} = \drawAngle{ABC,CBE}$, $\therefore \drawAngle{BCA} = \drawAngle{ABC}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop VI. Theor.}, reference=prop:I.VI]
\defineNewPicture[1/4]{
pair A, B, C, D;
A := (0, 0);
B := A shifted (7/2u, 0);
D := A shifted (7/4u, 3u);
C := 2/3[A, D];
draw byAngleWithName(B, A, C, byyellow, 0)(A);
draw byAngleWithName(A, B, D, black, 0)(B);
byLineDefine(B, C, byyellow, 0, 0);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(B, D, byblue, 0, 0);
byLineDefine(C, A, black, 0, 0);
byLineDefine(C, D, black, 1, 0);
draw byNamedLine(BC);
draw byNamedLineSeq(0)(CA,CD,BD,AB);
}
\drawCurrentPictureInMargin
\problemNP{I}{n}{any triangle (\drawLine[bottom][triangleABD]{CA,CD,BD,AB}) if two angles (\drawAngle{A} and \drawAngle{B}) are equal, the sides (\drawUnitLine{CA,CD} and \drawUnitLine{BD}) opposite to them are also equal.}
For if the sides be not equal, let one of them \drawUnitLine{CA,CD} be greater than the other \drawUnitLine{BD}, and from it to cut off $\drawUnitLine{CA} = \drawUnitLine{BD}$ \inprop[prop:I.III], draw \drawUnitLine{BC}.
\startCenterAlign
Then in \drawLine[bottom]{BC,AB,CA} and \triangleABD,\\
$\drawUnitLine{CA} = \drawUnitLine{BD}$ (const.),\\
$\drawAngle{A} = \drawAngle{B}$ (hyp.)\\
and \drawUnitLine{AB} common,\\
$\therefore$ the triangles are equal \inprop[prop:I.IV]\\
a part equal to the whole, which is absurd;\\
$\therefore$ neither of the sides \drawUnitLine{CA,CD} or \drawUnitLine{BD} is greater than the other,\\
$\therefore$ hence they are equal.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop VII. Theor.}, reference=prop:I.VII]
\defineNewPicture{
pair A, B, C, D, E, F, G, H;
A := (0, 0);
B := A shifted (4u, 0);
C := A shifted (u, 3u);
D := C shifted (7/4u, 0);
E := 1/2[C, D] yscaled -0.7;
F := E shifted (0, -2u);
G := 5/4[A, E];
H := 5/4[A, F];
draw byAngleWithName(B, C, A, black, 0)(C);
draw byAngle(D, C, B, byred, 0);
draw byAngleWithName(A, D, B, byyellow, 0)(D);
draw byAngle(C, D, A, byblue, 0);
draw byAngle(B, F, H, black, 0);
draw byAngle(B, F, E, byred, 0);
draw byAngle(B, E, G, byyellow, 0);
draw byAngle(G, E, F, byblue, 0);
draw byLine(C, D, black, 1, 0);
draw byLine(E, F, black, 1, 0);
draw byLine(A, B, black, 0, 0);
byLineDefine(B, C, byblue, 0, 0);
byLineDefine(C, A, byred, 0, 0);
byLineDefine(B, D, byblue, 0, 0);
byLineDefine(D, A, byred, 0, 0);
byLineDefine(B, E, byblue, 0, 0);
byLineDefine(E, A, byred, 0, 0);
byLineDefine(B, F, byblue, 0, 0);
byLineDefine(F, A, byred, 0, 0);
byLineDefine(E, G, byred, 1, 0);
byLineDefine(F, H, byred, 1, 0);
draw byNamedLine(EG,FH);
draw byNamedLineSeq(0)(BC,CA,EA,BE);
draw byNamedLineSeq(0)(BD,DA,FA,BF);
}
\drawCurrentPictureInMargin
\problemNP{O}{n}{the same base (\drawUnitLine{AB}), and on the same side of it there can not be two triangles having their conterminous sides (\drawUnitLine{CA} and \drawUnitLine{DA}, \drawUnitLine{BC} and \drawUnitLine{BD}) at both extremities of the base, equal to each other.}
When two triangles stand on the same base, and on the same side of it, the vertex of the one shall either fall outside of the other triangle, or within it; or, lastly, on one of its sides.
If it be possible let the two triangles be constructed so that $\left\{\eqalign{\drawUnitLine{CA}&=\drawUnitLine{BC}\cr \drawUnitLine{DA}&=\drawUnitLine{BD}\cr}\right\}$, then draw \drawUnitLine{CD} and,
\startCenterAlign
$\drawAngle{C,DCB} = \drawAngle{CDA}$ \inprop[prop:I.V]
$\therefore\drawAngle{DCB} < \drawAngle{CDA} \mbox{and}$
$\left.
\eqalign{
\therefore\drawAngle{DCB} &< \drawAngle{CDA,D}\cr
\mbox{but \inprop[prop:I.V]} \drawAngle{DCB} &= \drawAngle{CDA,D}
}\right\}\mbox{which is absurd,}$
\stopCenterAlign
therefore the two triangles cannot have their conterminous sides equal at both extremities of the base.
\qed
\stopProposition
\startProposition[title={Prop VIII. Theor.}, reference=prop:I.VIII]
\defineNewPicture{
pair A, B, C, D, E, F, d;
A := (0, 0);
B := A shifted (-u, -4u);
C := A shifted (3/2u, -3u);
d := (0, -9/2u);
D := A shifted d;
E := B shifted d;
F := C shifted d;
draw byAngleWithName(F, D, E, black, 0)(D);
draw byAngleWithName(C, A, B, black, 0)(A);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(B, C, black, 0, 0);
byLineDefine(C, A, byyellow, 0, 0);
byLineDefine(D, E, byred, 0, 1);
byLineDefine(E, F, black, 0, 1);
byLineDefine(F, D, byyellow, 0, 1);
draw byNamedLineSeq(0)(CA,BC,AB);
draw byNamedLineSeq(0)(FD,EF,DE);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{two triangles have two sides of the one respectively equal to two sides of the other ($\drawUnitLine{CA} = \drawUnitLine{FD}$ and $\drawUnitLine{AB} = \drawUnitLine{DE}$) and also their bases ($\drawUnitLine{BC} = \drawUnitLine{EF}$), equal; then the angles
(\drawFromCurrentPicture{
draw byNamedAngle(A) rotated -angleDirection.A;
} and
\drawFromCurrentPicture{
draw byNamedAngle(D) rotated -angleDirection.D;
}) contained by their equal sides are also equal.}
If the equal bases \drawUnitLine{BC} and \drawUnitLine{EF} be conceived to be placed one upon the other, so that the triangles shall lie at the same side of them, and that equal sides \drawUnitLine{AB} and \drawUnitLine{DE}, \drawUnitLine{CA} and \drawUnitLine{FD} be conterminous, the vertex of the one must fall on the vertex of the other; for to suppose them not coincident would contradict the last proposition.
Therefore sides \drawUnitLine{AB} and \drawUnitLine{CA}, being coincident with \drawUnitLine{DE} and \drawUnitLine{FD}, $\therefore \drawAngle{A} = \drawAngle{D}$.
\qed
\stopProposition
\startProposition[title={Prop IX. Prob.}, reference=prop:I.IX]
\defineNewPicture{
pair A, B, C, D, E, F;
A := (0, 2u);
B := (-4/3u, 0);
C := B xscaled -1;
D := A yscaled -1;
E := 5/4[A, B];
F := 5/4[A, C];
draw byAngle(B, A, D, byblue, 0);
draw byAngle(C, A, D, byyellow, 0);
byLineDefine(B, C, byyellow, 0, 0);
byLineDefine(A, D, black, 0, 0);
byLineDefine(D, B, byblue, 0, 0);
byLineDefine(C, D, byblue, 0, 0);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(C, A, byred, 0, 0);
byLineDefine(B, E, byred, 1, 0);
byLineDefine(C, F, byred, 1, 0);
draw byNamedLine(BC,AD);
draw byNamedLineSeq(0)(DB,CD);
draw byNamedLineSeq(0)(BE,AB,CA,CF);
}
\drawCurrentPictureInMargin
\problemNP{T}{o}{bisect a given rectilinear angle (\drawAngle{BAD,CAD}).}
\startCenterAlign
Take $\drawUnitLine{AB} = \drawUnitLine{CA}$ \inprop[prop:I.III]
draw \drawUnitLine{BC} upon which describe \drawLine{CD,DB,BC} \inprop[prop:I.I],\\
draw \drawUnitLine{AD}.
Because $\drawUnitLine{AB} = \drawUnitLine{CA}$ (const.)\\
and \drawUnitLine{AD} common to the two triangles\\
and $\drawUnitLine{CD} = \drawUnitLine{DB}$ (const.),
$\therefore \drawAngle{BAD} = \drawAngle{CAD}$ \inprop[prop:I.VIII].
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop X. Prob.}, reference=prop:I.X]
\defineNewPicture{
pair A, B, C, D;
A := (0, 3u);
B := (-7/4u, 0);
C := B xscaled -1;
D := 1/2[B, C];
draw byAngle(B, A, D, byblue, 0);
draw byAngle(C, A, D, byyellow, 0);
draw byLine(A, D, byred, 0, 0);
byLineDefine(D, B, black, 0, 0);
byLineDefine(C, D, black, 1, 0);
byLineDefine(A, B, byyellow, 0, 0);
byLineDefine(C, A, byblue, 0, 0);
draw byNamedLineSeq(0)(AB,CA,CD,DB);
}
\drawCurrentPictureInMargin
\problemNP{T}{o}{bisect a given finite straight line (\drawUnitLine{DB,CD}).}
\startCenterAlign
Construct \drawLine[bottom]{AB,CA,CD,DB} \inprop[prop:I.I],\\
draw \drawUnitLine{AD}, making $\drawAngle{BAD} = \drawAngle{CAD}$ \inprop[prop:I.IX],
Then $\drawUnitLine{DB} = \drawUnitLine{CD}$ by \inprop[prop:I.IV],
for~$\drawUnitLine{AB} = \drawUnitLine{CA}$ (const.) $\drawAngle{BAD} = \drawAngle{CAD}$\\
and \drawUnitLine{AD} common to the two triangles.
Therefore the given line is bisected.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XI. Prob.}, reference=prop:I.XI]
\defineNewPicture[1/4]{
pair A, B, C, D, E, F;
A := (0, 3u);
B := (-7/4u, 0);
C := B xscaled -1;
D := 1/2[B, C];
E := 3/2[D, B];
F := 3/2[D, C];
draw byAngle(A, D, B, byred, 0);
draw byAngle(C, D, A, byblue, 0);
draw byLine(A, D, byyellow, 0, 0);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(C, A, byblue, 0, 0);
draw byNamedLineSeq(0)(AB,CA);
draw byLine(D, B, black, 0, 0);
draw byLine(B, E, black, 1, 0);
draw byLine(C, D, byred, 0, 0);
draw byLine(F, C, byred, 1, 0);
}
\drawCurrentPictureInMargin
\problemNP{F}{rom}{a given point (\drawUnitLine{DB,CD}) in a given straight line (\drawUnitLine[3/2cm]{DB,CD}), to draw a perpendicular.}
\startCenterAlign
Take any point (\drawUnitLine{CD,FC}) in the given line,\\
cut off $\drawUnitLine{DB} = \drawUnitLine{CD}$ \inprop[prop:I.III],\\
construct \drawLine[bottom]{AB,CA,CD,DB} \inprop[prop:I.I],\\
draw \drawUnitLine{AD} and it shall be perpendicular to the given line.
for $\drawUnitLine{AB} = \drawUnitLine{CA}$ (const.)\\
$\drawUnitLine{CD} = \drawUnitLine{DB}$ (const.)\\
and \drawUnitLine{AD} common to the two triangles.
Therefore $\drawAngle{ADB} = \drawAngle{CDA}$ \inprop[prop:I.VIII]
$\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \indef[def:X]
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XII. Prob.}, reference=prop:I.XII]
\defineNewPicture{
pair A, B, C, D, E, F;
path c;
numeric r, a[];
A := (0, 3u);
B := (-7/4u, 0);
C := B xscaled -1;
D := 1/2[B, C];
E := 4/3[D, B];
F := 4/3[D, C];
r := arclength(A--B);
c := fullcircle scaled 2r shifted A;
a1 := xpart(c intersectiontimes (F--1/2[B, C]));
a2 := xpart(c intersectiontimes (E--1/2[B, C]));
draw byAngle(A, D, B, byyellow, 0);
draw byAngle(C, D, A, byblue, 0);
draw byLine(A, D, byred, 0, 0);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(C, A, byblue, 0, 0);
draw byNamedLineSeq(0)(AB,CA);
draw byArc(A, r, a1, a2, byred, 0, 0, 0, 0)(O);
draw byArc(A, r, a2-1/4, a2, byred, 1, 0, 0, 0)(Ol);
draw byArc(A, r, a1, a1+1/4, byred, 1, 0, 0, 0)(Or);
draw byLine(D, B, black, 0, 0);
draw byLine(B, E, black, 1, 0);
draw byLine(C, D, byyellow, 0, 0);
draw byLine(F, C, byyellow, 1, 0);
}
\drawCurrentPictureInMargin
\problemNP{T}{o}{draw a straight line perpendicular to a given indefinite straight line (\drawUnitLine[1.2cm]{DB,CD}) from a~given point (\drawFromCurrentPicture[middle][pointA]{
startTempScale(scaleFactor/2);
draw byNamedLine(AD);
draw byNamedLineSeq(0)(AB,CA);
stopTempScale;
}) without.}
\startCenterAlign
With the given point \pointA\ as centre, at one side of the line, and any distance \drawUnitLine{DB} capable of extending to the other side, describe \drawArc{O}.
Make $\drawUnitLine{DB} = \drawUnitLine{CD}$ \inprop[prop:I.X],\\
draw \drawUnitLine{AB}, \drawUnitLine{CA} and \drawUnitLine{AD}.
Then $\drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$.
For \inprop[prop:I.VIII] since $\drawUnitLine{DB} = \drawUnitLine{CD}$ (const.),\\
\drawUnitLine{AD} common to both,\\
and $\drawUnitLine{AB} = \drawUnitLine{CA}$ \indef[def:XV],
$\therefore \drawAngle{ADB} = \drawAngle{CDA}$, and
$\therefore \drawUnitLine{AD} \perp \drawUnitLine{DB,CD}$ \indef[def:X]
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XIII. Theor.}, reference=prop:I.XIII]
\defineNewPicture{
pair A, B, C, D, E;
A := (0, 5/2u);
B := (-7/4u, 0);
C := B xscaled -1;
D := (xpart(A), ypart(B));
E := (2/3xpart(C), 2/3ypart(A));
draw byAngle(A, D, B, byyellow, 0);
draw byAngle(E, D, A, byred, 0);
draw byAngle(C, D, E, byblue, 0);
draw byLine(A, D, black, 0, 0);
draw byLine(E, D, byyellow, 0, 0);
draw byLine(B, C, byred, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP{W}{hen}{a straight line (\drawUnitLine{ED}) standing upon another straight line (\drawUnitLine{BC}) makes angles with it; they are either two right angles or together equal to two right angles.}
\startCenterAlign
If \drawUnitLine{ED} be $\perp$ to \drawUnitLine{BC} then,\\
\drawAngle{ADB,EDA} and $\drawAngle{CDE} = \drawTwoRightAngles$ \indef[def:X],
but if \drawUnitLine{ED} be not $\perp$ to \drawUnitLine{BC},\\
draw $\drawUnitLine{AD} \perp \drawUnitLine{BC}$; \inprop[prop:I.XI]\\
$\drawAngle{ADB} +\drawAngle{CDE,EDA} = \drawTwoRightAngles$ (const.),\\
$\drawAngle{ADB} = \drawAngle{CDE,EDA} = \drawAngle{EDA} + \drawAngle{CDE}$
$\therefore \drawAngle{ADB} + \drawAngle{CDE,EDA} = \drawAngle{ADB} + \drawAngle{EDA} + \drawAngle{CDE}$ \inax[ax:II]
$= \drawAngle{ADB,EDA} + \drawAngle{CDE} = \drawTwoRightAngles$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XIV. Theor.}, reference=prop:I.XIV]
\defineNewPicture[1/4]{
pair A, B, C, D, E;
A := (u, 5/2u);
B := (-7/4u, 0);
C := B xscaled -1;
D := (0, 0);
E := (xpart(C), -1/2ypart(A));
draw byAngle(B, D, A, byyellow, 0);
draw byAngle(C, D, A, byblue, 0);
draw byAngle(E, D, C, byred, 0);
draw byLine(A, D, byred, 0, 0);
draw byLine(E, D, byyellow, 0, 0);
draw byLine(B, D, byblue, 0, 0);
draw byLine(C, D, black, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{two straight lines (\drawUnitLine{BD} and \drawUnitLine{CD}) meeting a third straight line (\drawUnitLine{AD}), at the same point, and at opposite sides of it, make with it adjacent angles (\offsetPicture{0pt}{15pt}{\drawAngle{BDA}} and \drawAngle{CDA}) equal to two right angles; these straight lines lie in one continuous straight line.}
\startCenterAlign
For, if possible let \drawUnitLine{ED}, and not \drawUnitLine{CD}, be the continuation of \drawUnitLine{BD},\\
then $\drawAngle{BDA} + \drawAngle{CDA,EDC} = \drawTwoRightAngles$
but by the hypothesis $\drawAngle{BDA} + \drawAngle{CDA} = \drawTwoRightAngles$
$\therefore\drawAngle{CDA,EDC} = \drawAngle{CDA}$, \inax[ax:III]; which is absurd \inax[ax:IX].
$\therefore \drawUnitLine{ED}$, is not the continuation of \drawUnitLine{BD}, and the like may be demonstrated of any other straight line except \drawUnitLine{CD}, $\therefore \drawUnitLine{CD}$ is the continuation of \drawUnitLine{BD}.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XV. Theor.}, reference=prop:I.XV]
\defineNewPicture{
pair A, B, C, D, E;
A := (7/4u, 3/2u);
B := A scaled -1;
C := A xscaled -1;
D := C scaled -1;
E := (A--B) intersectionpoint (C--D);
draw byAngle(B, E, C, byyellow, 0);
draw byAngle(C, E, A, byred, 0);
draw byAngle(A, E, D, black, 0);
draw byAngle(D, E, B, byblue, 0);
draw byLine(A, B, byred, 0, 0);
draw byLine(C, D, black, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{two straight lines (\drawUnitLine{AB} and \drawUnitLine{CD}) intersect on another, the vertical angles \drawAngle{BEC} and \drawAngle{AED}, \drawAngle{CEA} and \drawAngle{DEB} are equal.}
\startCenterAlign
$\drawAngle{BEC} + \drawAngle{CEA} = \drawTwoRightAngles$
$\drawAngle{AED} + \drawAngle{CEA} = \drawTwoRightAngles$
$\therefore \drawAngle{BEC} = \drawAngle{AED}$.
In the same manner it may be shown that\\
$\drawAngle{CEA} = \drawAngle{DEB}$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XVI. Theor.}, reference=prop:I.XVI]
\defineNewPicture[1/4]{
pair A, B, C, D, E, F, G;
A := (0, 0);
B := A shifted (3/2u, 7/2u);
C := A shifted (3u, 0);
D := B shifted (3u, 0);
E = whatever[A, D] = whatever[B, C];
F := (xpart(D), ypart(A));
G := 4/3[B, C];
draw byAngleWithName(B, A, C, byblue, 0)(A);
draw byAngleWithName(C, B, A, black, 0)(B);
draw byAngle(A, E, B, byyellow, 0);
draw byAngle(D, E, C, byyellow, 0);
draw byAngle(E, C, D, black, 0);
draw byAngle(G, C, A, byred, 0);
draw byAngle(D, C, F, black, 1);
byLineDefine(C, F, black, 1, 0);
byLineDefine(C, G, black, 0, 0);
byLineDefine(B, E, byblue, 0, 0);
byLineDefine(E, C, byblue, 1, 0);
byLineDefine(A, E, byred, 0, 0);
byLineDefine(E, D, byred, 1, 0);
byLineDefine(A, B, byyellow, 1, 0);
byLineDefine(A, C, black, 0, 0);
byLineDefine(C, D, byyellow, 0, 0);
draw byNamedLineSeq(0)(AE,ED,CD);
draw byNamedLineSeq(0)(EC,CG,noLine,CF,AC,AB,BE);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a side of a triangle (\drawLine[bottom]{BE,EC,AC,AB}) is produced, the external angle (\drawFromCurrentPicture[bottom][anglesECDpDCF]{
draw byNamedAngleSides(ECD,DCF)(CF);
}) is greater than either of the internal remote angles (\drawAngle{B} or \drawAngle{A})
}
\startCenterAlign
Make $\drawUnitLine{BE} = \drawUnitLine{EC}$ \inprop[prop:I.X];\\
Draw \drawUnitLine{AE} and produce it until $\drawUnitLine{ED} = \drawUnitLine{AE}$;\\
draw \drawUnitLine{CD}. In \drawLine{BE,AE,AB} and \drawLine{EC,ED,CD};\\
$\drawUnitLine{BE} = \drawUnitLine{EC}$, $\drawAngle{AEB} = \drawAngle{DEC}$ \inprop[prop:I.XV] and $\drawUnitLine{AE} = \drawUnitLine{ED}$ (const.),\\
$\therefore \drawAngle{B} = \drawAngle{ECD}$ \inprop[prop:I.IV],\\
$\therefore \anglesECDpDCF\ > \drawAngle{ECD}$.
In like manner it can be shown, that if \drawUnitLine{AC,CF} be produced, $\drawAngle{GCA} > \drawAngle{A}$ \\
and therefore \anglesECDpDCF\ which is $= \drawAngle{GCA}$ is $> \drawAngle{A}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XVII. Theor.}, reference=prop:I.XVII]
\defineNewPicture{
pair A, B, C, D;
A := (0, 0);
B := A shifted (3/2u, 5/2u);
C := A shifted (9/4u, 0);
D := C shifted (u, 0);
draw byAngleWithName(B, A, C, byblue, 0)(A);
draw byAngleWithName(A, B, C, black, 0)(B);
draw byAngle(A, C, B, byred, 0);
draw byAngle(B, C, D, byyellow, 0);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(B, C, byblue, 0, 0);
byLineDefine(A, C, black, 0, 0);
byLineDefine(C, D, black, 0, 0);
draw byNamedLineSeq(0)(noLine,BC,AB,AC,CD);
}
\drawCurrentPictureInMargin
\problemNP{A}{ny}{two angles of a triangle \drawLine[bottom]{AB,BC,AC} are together less than two right angles.}
\startCenterAlign
Produce \drawUnitLine{CD}, then will\\
$\drawAngle{ACB} + \drawAngle{BCD} = \drawTwoRightAngles$
But $\drawAngle{BCD} > \drawAngle{A}$ \inprop[prop:I.XVI]
$\therefore \drawAngle{ACB} + \drawAngle{A} < \drawTwoRightAngles$,
\stopCenterAlign
\noindent and in the same manner it may be shown that any other two angles of the triangle taken together are less than two right angles.
\qed
\stopProposition
\startProposition[title={Prop XVIII. Theor.}, reference=prop:I.XVIII]
\defineNewPicture[1/4]{
pair A, B, C, D;
A := (0, 0);
B := A shifted (5/2u, -1/2u);
C := A shifted (3/2u, 2u);
D := 2[C, A];
draw byAngleWithName(C, A, B, byblue, 0)(A);
draw byAngle(A, B, C, black, 0);
draw byAngle(D, B, A, byred, 0);
draw byAngleWithName(B, D, A, byyellow, 0)(D);
draw byLine(A, B, byyellow, 0, 0);
byLineDefine(A, C, byred, 0, 0);
byLineDefine(B, C, byblue, 0, 0);
byLineDefine(B, D, black, 0, 0);
byLineDefine(D, A, byred, 1, 0);
draw byNamedLineSeq(0)(AC,BC,BD,DA);
}
\drawCurrentPictureInMargin
\problemNP{I}{n}{any triangle \drawLine{AC,BC,BD,DA} if one side \drawUnitLine{DA,AC} be greater than another \drawUnitLine[0.5cm]{BC}, the angle opposite to the greater side is greater than the angle opposite to the less. i. e. $\drawAngle{ABC,DBA} > \drawAngle{D}$}
\startCenterAlign
Make $\drawUnitLine{AC} = \drawUnitLine{BC}$ \inprop[prop:I.III], draw \drawUnitLine{AB},
Then will $\drawAngle{A} = \drawAngle{ABC}$ \inprop[prop:I.V];
but $\drawAngle{ABC} > \drawAngle{D}$ \inprop[prop:I.XVI]
$\therefore \drawAngle{ABC} > \drawAngle{D}$ and much more\\
is $\drawAngle{ABC,DBA} > \drawAngle{D}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XIX. Theor.}, reference=prop:I.XIX]
\defineNewPicture[1/4]{
pair A, B, C;
A := (0, 0);
B := A shifted (7/2u, 0);
C := A shifted (u, 3u);
draw byAngleWithName(C, A, B, byblue, 0)(A);
draw byAngleWithName(A, B, C, byred, 0)(B);
byLineDefine(A, B, black, 0, 0);
byLineDefine(B, C, byblue, 0, 0);
byLineDefine(C, A, byred, 0, 0);
draw byNamedLineSeq(0)(CA,BC,AB);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{in any triangle \drawLine[bottom]{CA,BC,AB} one angle \drawAngle{A} be greater than another \drawAngle{B} the side \drawUnitLine{BC} which is opposite to the greater angle, is greater than the side \drawUnitLine{CA} opposite the less}
\startCenterAlign
If \drawUnitLine{BC} be not greater than \drawUnitLine{CA} then must\\
$\drawUnitLine{BC} =$ or $< \drawUnitLine{CA}$.
If $\drawUnitLine{BC} = \drawUnitLine{CA}$ then\\
$\drawAngle{A} = \drawAngle{B}$ \inprop[prop:I.V];\\
which is contrary to the hypothesis.
\drawUnitLine{BC} is not less than \drawUnitLine{CA}; for if it were,\\
$\drawAngle{A} < \drawAngle{B}$ \inprop[prop:I.XVIII]\\
which is contrary to the hypothesis:
$\therefore \drawUnitLine{BC} < \drawUnitLine{CA}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XX. Theor.}, reference=prop:I.XX]
\defineNewPicture{
pair A, B, C, D;
A := (0, 0);
B := A shifted (7/2u, 0);
D := A shifted (4/3u, 3/2u);
C := ((fullcircle scaled 2arclength(D--B)) shifted D) intersectionpoint (D--10[A, D]);
draw byAngleWithName(B, C, A, byred, 0)(C);
draw byAngle(C, B, D, byblue, 0);
draw byAngle(D, B, A, byyellow, 0);
byLineDefine(B, D, byred, 0, 0);
byLineDefine(A, B, black, 0, 0);
byLineDefine(B, C, byyellow, 0, 0);
byLineDefine(C, D, byblue, 1, 0);
byLineDefine(D, A, byblue, 0, 0);
draw byNamedLineSeq(0)(BD);
draw byNamedLineSeq(0)(DA,CD,BC,AB);
}
\drawCurrentPictureInMargin
\problemNP{A}{ny}{two sides \drawUnitLine{DA} and \drawUnitLine{BD} of a triangle \drawLine[bottom]{DA,BD,AB} taken together are greater than the third side (\drawUnitLine{AB}).}
\startCenterAlign
Produce \drawUnitLine{DA}, and\\
make $\drawUnitLine{CD} = \drawUnitLine{BD}$ \inprop[prop:I.III];\\
draw \drawUnitLine{BC}.
Then because $\drawUnitLine{CD} = \drawUnitLine{BD}$ (const.),\\
$\drawAngle{CBD} = \drawAngle{C}$ \inprop[prop:I.V]
$\therefore \drawAngle{CBD,DBA} > \drawAngle{C}$ \inax[ax:IX]
$\therefore \drawUnitLine{DA} + \drawUnitLine{CD} > \drawUnitLine{AB}$ \inprop[prop:I.XIX]
and $\therefore \drawUnitLine{DA} + \drawUnitLine{BD} > \drawUnitLine{AB}$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXI. Theor.}, reference=prop:I.XXI]
\defineNewPicture{
pair A, B, C, D, E;
A := (0, 0);
B := A shifted (7/2u, 0);
C := A shifted (3u, 4u);
D := 1/2[1/2[A, B], C];
E = whatever[A, D] = whatever[B, C];
draw byAngleWithName(B, D, A, byred, 0)(D);
draw byAngleWithName(B, E, D, byblue, 0)(E);
draw byAngleWithName(B, C, A, byyellow, 0)(C);
byLineDefine(B, D, byyellow, 0, 0);
byLineDefine(A, D, black, 0, 0);
byLineDefine(D, E, black, 1, 0);
byLineDefine(A, B, byblue, 1, 0);
byLineDefine(B, E, byred, 1, 0);
byLineDefine(E, C, byred, 0, 0);
byLineDefine(C, A, byblue, 0, 0);
draw byNamedLine(BD);
draw byNamedLineSeq(0)(AD,DE);
draw byNamedLineSeq(0)(CA,EC,BE,AB);
}
\drawCurrentPictureInMargin
\problemNP[2]{I}{f}{from any point (\drawFromCurrentPicture{
startTempScale(1/5);
draw byNamedLineSeq(0)(AD,BD);
stopTempScale;
}) within a triangle \drawFromCurrentPicture[bottom]{
startTempScale(1/5);
draw byNamedLineSeq(0)(CA,EC,BE,AB);
stopTempScale;
} straight lines be drawn to the extremities of one side (\drawSizedLine{DE}), these lines must be together less than the two other sides, but must contain a greater angle.}
\startCenterAlign
Produce \drawSizedLine{AD},\\
$\drawSizedLine{CA} + \drawSizedLine{EC} > \drawSizedLine{AD,DE}$ \inprop[prop:I.XX],\\
add \drawSizedLine{BE} to each,\\
$\drawSizedLine{CA} + \drawSizedLine{EC,BE} > \drawSizedLine{AD,DE} + \drawSizedLine{BE}$ \inax[ax:IV]
in the same manner it may be shown that\\
$\drawSizedLine{AD,DE} + \drawSizedLine{BE} > \drawSizedLine{AD} + \drawSizedLine{EC}$,\\
$\therefore \drawSizedLine{CA} + \drawSizedLine{EC,BE} > \drawSizedLine{AD} + \drawSizedLine{EC}$,\\
which was to be proved.
Again $\drawAngle{E} > \drawAngle{C}$ \inprop[prop:I.XVI],\\
and also $\drawAngle{D} > \drawAngle{E}$ \inprop[prop:I.XVI],
$\therefore \drawAngle{D} > \drawAngle{C}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXII. Theor.}, reference=prop:I.XXII]
\defineNewPicture[1/2]{
numeric r[], d;
pair A, B, C, D, E, LI, LII, LIII, LIV, LV, LVI;
path q[];
r1 := 3/2u;
r2 := 4/3u;
r3 := (2/3)*(r1+r2);
d := 1/4u;
A := (0, 0);
B := A shifted (r3, 0);
q1 := (fullcircle scaled 2r1) shifted A;
q2 := (fullcircle scaled 2r2) shifted B;
C := q1 intersectionpoint q2;
D := point 11/2 of q1;
E := point 3/4 of q2;
LI := (xpart(point 0 of q2), ypart(point 6 of q1));
LII := LI shifted (-r3, 0);
LIII := LI shifted (0, -d);
LIV := LIII shifted (-r2, 0);
LV := LIII shifted (0, -d);
LVI := LV shifted (-r1, 0);
draw byCircle(A, r1, byblue, 0, 0, 0)(A);
byLineDefine(A, D, byblue, 0, 0);
byLineDefine(B, E, byred, 0, 0);
byLineDefine(A, B, black, 0, 0);
byLineDefine(B, C, byyellow, 0, 0);
byLineDefine(C, A, byyellow, 1, 0);
draw byNamedLineSeq(0)(BC,CA);
draw byNamedLineSeq(0)(AD,AB,BE);
draw byLineWithName (LI, LII, black, 1, 0)(Lu);
draw byLineWithName (LIII, LIV, byred, 1, 0)(Ld);
draw byLineWithName (LV, LVI, byblue, 1, 0)(Lt);
draw byCircle(B, r2, byred, 0, 0, 0)(B);
}
\drawCurrentPictureInMargin
\problemNP{G}{iven}{three right lines $\left\{\vcenter{
\nointerlineskip\hbox{\drawSizedLine{Lu}}
\nointerlineskip\hbox{\drawSizedLine{Ld}}
\nointerlineskip\hbox{\drawSizedLine{Lt}}}\right.$
the sum of any two greater than the third, to construct a triangle whose sides shall be respectively equal to the given lines.}
\startCenterAlign
Assume $\drawSizedLine{AB} = \drawSizedLine{Lu}$ \inprop[prop:I.III].
$\left.\eqalign{
\mbox{Draw } \drawSizedLine{AD} &= \drawSizedLine{Ld}\cr
\mbox{and } \drawSizedLine{BE} &= \drawSizedLine{Lt}
}\right\}\mbox{\inprop[prop:I.II].}$
With \drawSizedLine{AD} and \drawSizedLine{BE} as radii, describe
\drawFromCurrentPicture{
draw byNamedLine (AD); draw byNamedCircle (A);
} and
\offsetPicture{12pt}{0pt}{\drawFromCurrentPicture{
draw byNamedLine (BE); draw byNamedCircle (B);
}} \inpost[post:III];\\
draw \drawSizedLine{CA} and \drawSizedLine{BC},\\
then will \drawLine[bottom]{CA,BC,AB} be the triangle required.
$\left.\eqalign{
\mbox{For } \drawSizedLine{AB} &= \drawSizedLine{Lu} \mbox{,} \cr
\drawSizedLine{BC} &= \drawSizedLine{BE} = \drawSizedLine{Ld} \cr
\mbox{and } \drawSizedLine{CA} &= \drawSizedLine{AD} = \drawSizedLine{Lt} \cr
}\right\}\mbox{(const.)}$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXIII. Prob.}, reference=prop:I.XXIII]
\defineNewPicture{
angleScale := 3/2;
pair A, B, C, D, E, F, G, H, J, d;
A := (0, 0);
B := A shifted (7/2u, 0);
C := A shifted (3u, 11/5u);
D := 5/4[A, B];
E := 7/6[A, C];
d := (0, -5/2u);
F := A shifted d;
G := B shifted d;
H := C shifted d;
J := D shifted d;
draw byAngleWithName(B, A, C, byred, 0)(A);
draw byAngleWithName(G, F, H, byblue, 0)(F);
byLineDefine(B, D, black, 1, 1);
byLineDefine(C, E, byblue, 1, 1);
byLineDefine(A, B, black, 0, 1);
byLineDefine(C, A, byblue, 0, 1);
draw byLine(B, C, byred, 0, 1);
draw byNamedLineSeq(0)(CE,CA,AB,BD);
byLineDefine(G, J, black, 1, 0);
byLineDefine(F, G, black, 0, 0);
byLineDefine(G, H, byred, 0, 0);
byLineDefine(H, F, byyellow, 0, 0);
draw byNamedLineSeq(0)(noLine,GH,HF,FG,GJ);
}
\drawCurrentPictureInMargin
\problemNP{A}{t}{a given point (\drawFromCurrentPicture{
startTempScale(scaleFactor/2);
draw byNamedLineSeq(0)(FG,HF);
stopTempScale;
}) in a given straight line (\drawUnitLine{FG,GJ}), to make an angle equal to a given rectilinear angle (\drawAngle{A}).}
Draw \drawUnitLine{BC} between any two points in the legs of the given angle.
\startCenterAlign
Construct \drawLine[bottom]{HF,GH,FG} \inprop[prop:I.XXII]\\
so that $\drawUnitLine{FG} = \drawUnitLine{AB}$, $\drawUnitLine{HF} = \drawUnitLine{CA}$\\
and $\drawUnitLine{GH} = \drawUnitLine{BC}$.
Then $\drawAngle{A} = \drawAngle{F}$ \inprop[prop:I.VIII].
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXIV. Theor.}, reference=prop:I.XXIV]
\defineNewPicture{
pair A, B, C, D, E, F, G, d;
A := (0, 0);
B := A shifted (u, -5/2u);
C := A shifted (-u, -7/2u);
D := (xpart(C) - 3/2u, ypart(B));
d := (0, -4u);
E := A shifted d;
F := B shifted d;
G := C shifted d;
draw byAngle(B, A, C, black, 2);
draw byAngle(C, A, D, black, 3);
draw byAngle(F, E, G, black, 2);
draw byAngle(B, D, A, byblue, 0);
draw byAngle(C, D, B, byred, 0);
draw byAngle(D, C, A, byyellow, 0);
draw byAngle(A, C, B, black, 0);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(B, C, black, 1, 0);
draw byLine(C, A, byred, 0, 0);
draw byLine(B, D, black, 0, 0);
byLineDefine(A, D, byred, 1, 0);
byLineDefine(C, D, byblue, 1, 0);
draw byNamedLineSeq(0)(AB,AD,CD,BC);
byLineDefine(E, F, byblue, 0, 1);
byLineDefine(F, G, byyellow, 0, 1);
byLineDefine(G, E, byred, 0, 1);
draw byNamedLineSeq(0)(EF,FG,GE);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{two triangles have two sides of the one respectively equal to two sides of the other (\drawUnitLine{AB} to \drawUnitLine{EF} and \drawUnitLine{AD} to \drawUnitLine{GE}), and if one of the angles
(\drawFromCurrentPicture[bottom]{
draw byNamedAngleSides(BAC,CAD)(AB,CA,AD);
})
contained by the equal sides be greater than the other
(\drawFromCurrentPicture[bottom][angleBAC]{
draw byNamedAngleSides(BAC)(AB, CA);
}),
the side (\drawUnitLine{BD}) which is opposite to the greater angle is greater than the side (\drawUnitLine{FG}) which is opposite to the less angle.
}
\startCenterAlign
Make $\angleBAC = \drawFromCurrentPicture[bottom][angleFEG]{
draw byNamedAngleSides(FEG)(EF, GE);
}$ \inprop[prop:I.XXIII],\\
and $\drawUnitLine{CA} = \drawUnitLine{GE}$ \inprop[prop:I.III],\\
draw \drawUnitLine{CD} and \drawUnitLine{BC}.
Because $\drawUnitLine{CA} = \drawUnitLine{AD}$ (\inaxL[ax:I]. hyp. const.)\\
$\therefore \drawAngle{BDA,CDB} = \drawAngle{DCA}$ \inprop[prop:I.V]
but $\drawAngle{CDB} < \drawAngle{DCA}$,\\
and $\therefore \drawAngle{CDB} < \drawAngle{DCA,ACB}$,
$\therefore \drawUnitLine{BD} > \drawUnitLine{BC}$ \inprop[prop:I.XIX]
but $\drawUnitLine{BC} = \drawUnitLine{FG}$ \inprop[prop:I.IV]
$\therefore \drawUnitLine{BD} > \drawUnitLine{FG}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXV. Theor.}, reference=prop:I.XXV]
\defineNewPicture{
pair A, B, C, D, E, F, d;
A := (0, 0);
B := A shifted (u, -3u);
C := A shifted (-7/4u, -4u);
d := (0, -9/2u);
D := A shifted d;
E := ((B shifted -A) rotated -10) shifted d;
F := C shifted d;
draw byAngleWithName(B, A, C, byyellow, 0)(A);
draw byAngleWithName(E, D, F, black, 0)(D);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(B, C, black, 0, 0);
byLineDefine(C, A, byred, 0, 0);
draw byNamedLineSeq(0)(AB,BC,CA);
byLineDefine(D, E, byblue, 0, 1);
byLineDefine(E, F, byyellow, 0, 1);
byLineDefine(F, D, byred, 0, 1);
draw byNamedLineSeq(0)(DE,EF,FD);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{two triangles have two sides (\drawUnitLine[0.7cm]{AB} and \drawUnitLine[0.7cm]{CA}) respectively equal to two sides (\drawUnitLine{DE} and \drawUnitLine{FD}) of the other, but their bases unequal, the angle subtended by the greater base (\drawUnitLine{BC}) of the one, must be greater than the angle subtended by the less base (\drawUnitLine{EF}) of the other.}
\startCenterAlign
$\drawAngle{A} =\mbox{, } > \mbox{ or } < \drawAngle{D}$
\drawAngle{A} is not equal to \drawAngle{D}\\
for if $\drawAngle{A} = \drawAngle{D}$ then $\drawUnitLine{BC} = \drawUnitLine{EF}$ \inprop[prop:I.IV]\\
which is contrary to the hypothesis;
\drawAngle{A} is not less than \drawAngle{D}\\
for if $\drawAngle{A} < \drawAngle{D}$\\
then $\drawUnitLine{BC} < \drawUnitLine{EF}$ \inprop[prop:I.XXIV],\\
which is also contrary to the hypothesis:
$\therefore \drawAngle{A} > \drawAngle{D}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXVI. Theor.}, reference=prop:I.XXVI]
\defineNewPicture{
pair A, B, C, D, E, F, G, d;
A := (0, 0);
B := A shifted (3u, 0);
C := A shifted (2u, 3u);
d := (0, -4u);
D := A shifted d;
E := B shifted d;
F := C shifted d;
G := 3/4[D, F];
draw byAngleWithName(B, A, C, byyellow, 0)(A);
draw byAngleWithName(C, B, A, byred, 0)(B);
draw byAngleWithName(A, C, B, black, 1)(C);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(B, C, black, 0, 0);
byLineDefine(C, A, byred, 0, 0);
draw byNamedLineSeq(0)(CA,BC,AB);
draw byAngleWithName(E, D, F, byyellow, 0)(D);
draw byAngle(G, E, D, black, 0);
draw byAngle(F, E, G, byblue, 0);
draw byAngleWithName(D, F, E, black, 1)(F);
draw byLine(E, G, byyellow, 0, 1);
byLineDefine(D, E, byblue, 0, 1);
byLineDefine(E, F, black, 0, 1);
byLineDefine(F, G, byred, 1, 1);
byLineDefine(G, D, byred, 0, 1);
draw byNamedLineSeq(0)(GD,FG,EF,DE);
}
\problemNP{I}{f}{two triangles have two angles of the one respectively equal to two angles of the other ($\drawAngle{A} = \drawAngle{D}$ and $\drawAngle{B} = \drawAngle{GED,FEG}$), and a side of the one equal to a side of the other similarly placed with respect to the equal angles, the remaining sides and angles are respectively equal to one another.}
\drawCurrentPictureInMargin
\startsubproposition[title={Case I.}]
\startCenterAlign
Let \drawUnitLine{AB} and \drawUnitLine{DE} which lie between the equal angles be equal,\\
then $\drawUnitLine{CA} = \drawUnitLine{GD,FG}$.
For if it be possible, let one of them \drawUnitLine{GD,FG} be greater than the other;\\
make $\drawUnitLine{CA} = \drawUnitLine{GD}$, draw \drawUnitLine{EG}.
In \drawLine[bottom]{CA,BC,AB} and
\drawLine[bottom]{GD,EG,DE} we have\\
$\drawUnitLine{AB} = \drawUnitLine{GD}$, $\drawAngle{A} = \drawAngle{D}$, $\drawUnitLine{AB} = \drawUnitLine{DE}$;\\
$\therefore \drawAngle{B} = \drawAngle{GED}$ (pr. 4.)\\
but $\drawAngle{B} = \drawAngle{GED,FEG}$ (hyp.)
and therefore $\drawAngle{GED} = \drawAngle{GED,FEG}$, which is absurd; hence neither of the sides \drawUnitLine{CA} and \drawUnitLine{GD,FG} is greater than the other; and $\therefore$ they are equal;
$\therefore \drawUnitLine{BC} = \drawUnitLine{EF}$, and $\drawAngle{C} = \drawAngle{F}$, \inprop[prop:I.IV].
\stopCenterAlign
\stopsubproposition
\vfill\pagebreak
\defineNewPicture{
pair H, I, J, K, L, M, N, d;
d := (0, -4u);
H := (0, 0);
I := H shifted (3u, 0);
J := H shifted (1u, 3u);
K := H shifted d;
L := I shifted d;
M := J shifted d;
N := 3/4[K, L];
draw byAngleWithName(I, H, J, byyellow, 0)(H);
draw byAngleWithName(J, I, H, byred, 0)(I);
byLineDefine(H, I, byblue, 0, 0);
byLineDefine(I, J, black, 0, 0);
byLineDefine(J, H, byred, 0, 0);
draw byNamedLineSeq(0)(JH,IJ,HI);
draw byAngleWithName(L, K, M, byyellow, 0)(K);
draw byAngleWithName(M, N, K, black, 0)(N);
draw byAngleWithName(M, L, K, byred, 0)(L);
draw byLine(M, N, byyellow, 0, 1);
byLineDefine(K, N, byblue, 0, 1);
byLineDefine(N, L, byblue, 1, 1);
byLineDefine(L, M, black, 0, 1);
byLineDefine(M, K, byred, 0, 1);
draw byNamedLineSeq(0)(MK,LM,NL,KN);
}
\drawCurrentPictureInMargin
\startsubproposition[title={Case II.}]
\startCenterAlign
Again, let $\drawUnitLine{JH} = \drawUnitLine{MK}$, which lie opposite the equal angles \drawAngle{I} and \drawAngle{L}. If it be possible, let $\drawUnitLine{KN,NL} > \drawUnitLine{HI}$, then take $\drawUnitLine{KN} = \drawUnitLine{HI}$, draw \drawUnitLine{MN}.
Then in \drawLine[bottom]{JH,IJ,HI} and \drawLine[bottom]{MK,MN,KN} we have $\drawUnitLine{JH} = \drawUnitLine{MK}$, $\drawUnitLine{HI} = \drawUnitLine{KN}$ and $\drawAngle{H} = \drawAngle{K}$,
$\therefore \drawAngle{I} = \drawAngle{N}$ \inprop[prop:I.IV]\\
but $\drawAngle{I} = \drawAngle{L}$ (hyp.)
$\therefore \drawAngle{N} = \drawAngle{L}$ which is absurd \inprop[prop:I.XVI]
Consequently, neither of the sides \drawUnitLine{HI} or \drawUnitLine{KN,NL} is greater than the other, hence they must be equal. It follows (by \inpropL[prop:I.IV]) that the triangles are equal in all respects.
\stopCenterAlign
\stopsubproposition
\qed
\stopProposition
\startProposition[title={Prop XXVII. Theor.}, reference=prop:I.XXVII]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, I, d;
A := (0, 0);
B := A shifted (8/3u, 0);
d := (0, -7/4u);
C := A shifted d;
D := B shifted d;
E := 1/3[A, B];
F := 2/3[C, D];
G := 3/2[F, E];
H := 3/2[E, F];
I := 1/2[A, C] shifted (-2u, 0);
draw byAngle(A, E, F, byyellow, 0);
draw byAngle(F, E, B, byred, 0);
draw byAngle(C, F, E, byblue, 0);
draw byAngle(E, F, D, byyellow, 0);
byLineDefine(I, A, byblue, 0, 0);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(I, C, byred, 0, 0);
byLineDefine(C, D, byred, 0, 0);
draw byNamedLineSeq(0)(CD,IC,IA,AB);
draw byLine(G, H, black, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a straight line (\drawUnitLine{GH}) meeting two other straight lines (\drawUnitLine{CD} and \drawUnitLine{AB}) makes with them the alternate angles (\drawAngle{CFE} and \drawAngle{FEB}; \drawAngle{EFD} and \drawAngle{AEF}) equal, these two straight lines are parallel.}
If \drawUnitLine{CD} be not parallel to \drawUnitLine{AB} they shall meet when produced.
If it be possible, let those lines be not parallel, but meet when produced; then the external angle \drawAngle{FEB} is greater than \drawAngle{CFE} \inprop[prop:I.XVI], but they are also equal (hyp.), which is absurd: in the same manner it may be shown that they cannot meet on the other side; $\therefore$ they are parallel.
\qed
\stopProposition
\startProposition[title={Prop XXVIII. Theor.}, reference=prop:I.XXVIII]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, d;
A := (0, 0);
B := A shifted (7/2u, 0);
d := (0, -3/2u);
C := A shifted d;
D := B shifted d;
E := 9/20[A, B];
F := 11/20[C, D];
G := 7/4[F, E];
H := 4/3[E, F];
draw byAngle(G, E, A, black, 0);
draw byAngle(B, E, G, byyellow, 0);
draw byAngle(A, E, F, byred, 0);
draw byAngle(F, E, B, byblue, 0);
draw byAngle(C, F, E, byblue, 0);
draw byAngle(E, F, D, byred, 0);
draw byLine(A, B, byred, 0, 0);
draw byLine(C, D, byyellow, 0, 0);
draw byLine(G, H, black, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a straight line (\drawUnitLine{GH}) meeting two other straight lines (\drawUnitLine{AB} and \drawUnitLine{CD}) makes the external equal to the internal and opposite angle, at the same side of the cutting line (namely $\drawAngle{GEA} = \drawAngle{CFE}$ or $\drawAngle{BEG} = \drawAngle{EFD}$), or if it makes two internal angles at the same side (\drawAngle{EFD} and \drawAngle{FEB}, or \drawAngle{CFE} and \drawAngle{AEF}) together equal to two right angles, those two straight lines are parallel.}
\startCenterAlign
First, if $\drawAngle{GEA} = \drawAngle{CFE}$, then $\drawAngle{GEA} = \drawAngle{FEB}$ \inprop[prop:I.XV],\\
$\therefore \drawAngle{CFE} = \drawAngle{FEB} \therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$ \inprop[prop:I.XXVII].
Secondly, if $\drawAngle{CFE} + \drawAngle{AEF} = \drawTwoRightAngles$,\\
then $\drawAngle{AEF} + \drawAngle{FEB} = \drawTwoRightAngles$ \inprop[prop:I.XIII],\\
$\therefore \drawAngle{CFE} + \drawAngle{AEF} = \drawAngle{AEF} + \drawAngle{FEB}$ \inax[ax:III]
$\therefore \drawAngle{CFE} = \drawAngle{FEB}$
$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXIX. Theor.}, reference=prop:I.XXIX]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, I, J, d[];
A := (0, 0);
B := A shifted (7/2u, 0);
d1 := (0, -2u);
C := A shifted d1;
D := B shifted d1;
E := 11/20[A, B];
F := 9/20[C, D];
G := 7/4[F, E];
H := 4/3[E, F];
d2 := (3/2u, 1/2u);
I := E shifted -d2;
J := E shifted d2;
draw byAngle(I, E, A, byblue, 0);
draw byAngle(F, E, I, byyellow, 0);
draw byAngle(B, E, F, black, 0);
draw byAngle(G, E, B, byred, 0);
draw byAngle(E, F, D, black, 0);
draw byLine(I, E, black, 0, 0);
draw byLine(E, J, black, 1, 0);
draw byLine(A, B, byyellow, 0, 0);
draw byLine(C, D, byred, 0, 0);
draw byLine(G, H, byblue, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP{A}{ straight}{line (\drawUnitLine{GH}) falling on two parallel straight lines (\drawUnitLine{AB} and \drawUnitLine{CD}), makes the alternate angles equal to one another; and also the external equal to the internal and opposite angle on the same side; and the two internal angles on the same side together equal to two right angles.}
For if the alternate angles \drawAngle{IEA,FEI} and \drawAngle{EFD} be not equal, draw \drawUnitLine{IE}, making $\drawAngle{FEI} = \drawAngle{EFD}$ \inprop[prop:I.XXIII].
Therefore $\drawUnitLine{IE,EJ} \parallel \drawUnitLine{CD}$ \inprop[prop:I.XXVII] and therefore two straight lines which intersect are parallel to the same straight line, which is impossible \inax[ax:XII].
Hence \drawAngle{IEA,FEI} and \drawAngle{EFD} are not unequal, that is, they are equal: $\drawAngle{IEA,FEI} = \drawAngle{GEB}$ \inprop[prop:I.XV]; $\therefore \drawAngle{GEB} = \drawAngle{EFD}$, the external angle equal to the internal and opposite on the same side: if \drawAngle{BEF} be added to both, then $\drawAngle{EFD} + \drawAngle{BEF} = \drawAngle{BEF,GEB} = \drawTwoRightAngles$ \inprop[prop:I.XIII]. That is to say, the two internal angles at the same side of the cutting line are equal to two right angles.
\qed
\stopProposition
\startProposition[title={Prop XXX. Theor.}, reference=prop:I.XXX]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, I, J, K, d;
A := (0, 0);
B := A shifted (7/2u, 0);
d := (0, -u);
C := A shifted d;
D := B shifted d;
E := C shifted d;
F := D shifted d;
G := 13/20[A, B];
H := 7/20[E, F];
I := (G--H) intersectionpoint (C--D);
J := 3/2[H, G];
K := 5/4[G, H];
draw byAngleWithName(B, G, J, byyellow, 0)(G);
draw byAngleWithName(D, I, J, byblue, 0)(I);
draw byAngleWithName(F, H, J, byred, 0)(H);
draw byLine(A, B, byred, 0, 0);
draw byLine(C, D, byyellow, 0, 0);
draw byLine(E, F, byblue, 0, 0);
draw byLine(J, K, black, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP{S}{traight}{lines (\drawUnitLine{AB} and \drawUnitLine{EF}) which are parallel to the same straight line (\drawUnitLine{CD}), are parallel to one another.}
\startCenterAlign
Let \drawUnitLine{JK} intersect $\left\{\vcenter{\nointerlineskip\hbox{\drawUnitLine{AB}}\nointerlineskip\hbox{\drawUnitLine{CD}}\nointerlineskip\hbox{\drawUnitLine{EF}}}\right\}$;
Then, $\drawAngle{G} = \drawAngle{I} = \drawAngle{H}$ \inprop[prop:I.XXIX],
$\therefore \drawAngle{G} = \drawAngle{H}$
$\therefore \drawUnitLine{AB} \parallel \drawUnitLine{EF}$ \inprop[prop:I.XXVII].
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXXI. Prob.}, reference=prop:I.XXXI]
\defineNewPicture[1/6]{
pair A, B, C, D, E, F, d;
A := (0, 0);
B := A shifted (7/2u, 0);
d := (0, -3u);
C := A shifted d;
D := B shifted d;
E := 4/5[A, B];
F := 1/5[C, D];
draw byAngleWithName(F, E, A, byyellow, 0)(E);
draw byAngleWithName(E, F, D, byred, 0)(F);
draw byLine(E, F, black, 0, 0);
draw byLine(A, E, byred, 0, 0);
draw byLine(E, B, byred, 1, 0);
draw byLine(C, F, byblue, 0, 0);
draw byLine(F, D, byblue, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP{F}{rom}{a given point \drawLine[middle][pointE]{AE,EF} to draw a straight line parallel to a given straight line (\drawUnitLine{CF}).}
\startCenterAlign
Draw \drawUnitLine{EF} from the point \pointE to any point \drawLine[middle][pointF]{FD,EF} in \drawUnitLine{CF},
make $\drawAngle{E} = \drawAngle{F}$ \inprop[prop:I.XXIII],
then $\drawUnitLine{AE,EB} \parallel \drawUnitLine{CF}$ \inprop[prop:I.XXVII].
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXXII. Theor.}, reference=prop:I.XXXII]
\defineNewPicture[1/6]{
pair A, B, C, D, E;
A := (0, 0);
B := A shifted (-7/4u, -3u);
C := A shifted (7/4u, -3u);
D := 4/3[B, A];
E := A shifted (unitvector(C-B) scaled 3/2u);
draw byAngle(D, A, E, byred, 0);
draw byAngle(E, A, C, black, 0);
draw byAngle(C, A, B, byblue, 0);
draw byAngleWithName(A, B, C, byyellow, 0)(B);
draw byAngleWithName(B, C, A, black, 0)(C);
draw byLine(A, E, byblue, 0, 0);
byLineDefine(A, D, black, 1, 0);
byLineDefine(A, B, black, 0, 0);
byLineDefine(B, C, byred, 0, 0);
byLineDefine(C, A, byyellow, 0, 0);
draw byNamedLineSeq(0)(CA,noLine,AD,AB,BC);
}
\drawCurrentPictureInMargin
\problemNP[2]{I}{f}{any side (\drawUnitLine{AB}) of a triangle by produced, the external angle (\drawAngle{DAE,EAC}) is equal to the sum of two internal and opposite angles (\drawAngle{B} and \drawAngle{C}), and the three internal angles of any triangle taken together are equal to two right angles.}
\startCenterAlign
Through the point \drawLine{AB,CA} draw\\
$\drawUnitLine{AE} \parallel \drawUnitLine{BC}$ \inprop[prop:I.XXXI].
Then $\left\{\eqalign{\drawAngle{DAE} &= \drawAngle{B}\cr \drawAngle{EAC} &= \drawAngle{C}\cr}\right\}$ \inprop[prop:I.XXIX],
$\therefore \drawAngle{B} + \drawAngle{C} = \drawAngle{DAE,EAC}$ \inax[ax:II],
and therefore\\
$\drawAngle{B} + \drawAngle{CAB} + \drawAngle{C} =
\drawFromCurrentPicture{
draw byNamedAngle(DAE);
draw byNamedAngle(EAC);
draw byNamedAngle(CAB);
} = \drawTwoRightAngles$ \inprop[prop:I.XIII]
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXXIII. Theor.}, reference=prop:I.XXXIII]
\defineNewPicture[1/4]{
pair A, B, C, D, d[];
d1 := (5/2u, 0);
d2 := (-7/8u, -3u);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
draw byAngle(B, A, D, byyellow, 0);
draw byAngle(D, A, C, byred, 0);
draw byAngle(C, D, A, black, 0);
draw byAngle(A, D, B, byblue, 0);
draw byLine(A, D, black, 0, 0);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(C, D, byred, 1, 0);
byLineDefine(A, C, byblue, 0, 0);
byLineDefine(B, D, byyellow, 0, 0);
draw byNamedLineSeq(0)(AB,BD,CD,AC);
}
\drawCurrentPictureInMargin
\problemNP{S}{traight}{lines (\drawUnitLine{AC} and \drawUnitLine{BD}) which join the adjacent extremities of two equal and parallel straight lines (\drawUnitLine{AB} and \drawUnitLine{CD}), are themselves parallel.}
\startCenterAlign
Draw \drawUnitLine{AD} the diagonal.
$\drawUnitLine{AB} = \drawUnitLine{CD}$ (hyp.)\\
$\drawAngle{BAD} = \drawAngle{CDA}$ \inprop[prop:I.XXIX]\\
and \drawUnitLine{AD} common to the two triangles;
$\therefore \drawUnitLine{AC} = \drawUnitLine{BD}$, and $\drawAngle{ADB} = \drawAngle{DAC}$ \inprop[prop:I.IV];
and $\therefore \drawUnitLine{AC} \parallel \drawUnitLine{BD}$ \inprop[prop:I.XXVII].
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXXIV. Theor.}, reference=prop:I.XXXIV]
\defineNewPicture{
pair A, B, C, D, d[];
d1 := (5/2u, 0);
d2 := (-7/8u, -3u);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
draw byAngle(B, A, D, byblue, 0);
draw byAngle(D, A, C, byred, 0);
draw byAngle(C, D, A, byyellow, 0);
draw byAngle(A, D, B, byred, 0);
draw byAngleWithName(A, C, D, black, 0)(C);
draw byAngleWithName(D, B, A, black, 0)(B);
draw byLine(A, D, black, 0, 0);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(C, D, byred, 1, 0);
byLineDefine(A, C, byyellow, 0, 0);
byLineDefine(B, D, byblue, 0, 0);
draw byNamedLineSeq(0)(AB,BD,CD,AC);
}
\drawCurrentPictureInMargin
\problemNP{T}{he}{opposite sides and angles of any parallelogram are equal, and the diagonal (\drawUnitLine{AD}) divides it into two equal parts.}
\startCenterAlign
Since $\left\{\eqalign{\drawAngle{BAD} &= \drawAngle{CDA}\cr\drawAngle{DAC} &= \drawAngle{ADB}\cr}\right\}$ \inprop[prop:I.XXIX]\\
and \drawUnitLine{AD} common to the two triangles.
$\therefore \left\{\eqalign{\drawUnitLine{AB} &= \drawUnitLine{CD}\cr \drawUnitLine{AC} &= \drawUnitLine{BD}\cr \drawAngle{B} &= \drawAngle{C}\cr}\right\}$ \inprop[prop:I.XXVI]\\
and $\drawAngle{BAD,DAC} = \drawAngle{CDA,ADB}$ \inax[ax:II]:
\stopCenterAlign
Therefore the opposite sides and angles of the parallelograms are equal: and as the triangles \drawLine{AD,CD,AC} and \drawLine{AB,BD,AD} are equal in every respect \inprop[prop:I.IV], the diagonal divides the parallelogram into two equal parts.
\qed
\stopProposition
\startProposition[title={Prop XXXV. Theor.}, reference=prop:I.XXXV]
\defineNewPicture{
pair A, B, C, D, E, F, G, d[];
d1 := (7/4u, 0);
d2 := (u, -3u);
d3 := (-2u, -3u);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
E := C shifted -d3;
F := D shifted -d3;
G := (B--D) intersectionpoint (C--E);
draw byPolygon(A,B,G,C)(byyellow);
draw byPolygon(E,F,D,G)(byyellow);
draw byPolygon(B,E,G)(byyellow);
draw byPolygon(C,D,G)(byblue);
draw byAngleWithName(B, A, C, byred, 0)(A);
draw byAngleWithName(E, B, D, byblue, 0)(B);
draw byAngleWithName(A, E, C, black, 0)(E);
draw byAngleWithName(A, F, D, white, 0)(F);
byLineDefine(A, C, byblue, 0, 0);
byLineDefine(B, D, byred, 0, 0);
byLineDefine(C, D, black, 0, 0);
draw byNamedLineSeq(0)(AC,CD,BD);
}
\drawCurrentPictureInMargin
\problemNP{P}{arallelograms}{on the same base, and between the same parallels, are (in area) equal.}
\startCenterAlign
On account of the parallels,\\
$\left.\eqalign{
\drawAngle{A} &= \drawAngle{B};\cr
\drawAngle{E} &= \drawAngle{F};\cr
\mbox{and } \drawUnitLine{AC} &= \drawUnitLine{BD}\cr
}\right\}
\eqalign{
&\mbox{\inprop[prop:I.XXIX]}\cr
&\mbox{\inprop[prop:I.XXIX]}\cr
&\mbox{\inprop[prop:I.XXXIV]}\cr
}$
But
$\drawFromCurrentPicture[middle][polygonABC]{
draw byNamedPolygon (ABGC, BEG);
draw byNamedLine (AC);
}
=
\drawFromCurrentPicture[middle][polygonEFD]{
draw byNamedPolygon (EFDG, BEG);
draw byNamedLine (BD);
}$ \inprop[prop:I.VIII]
$\therefore
\drawFromCurrentPicture[middle][polygonAFDC]{
draw byNamedPolygon (ABGC, EFDG, BEG, CDG);
draw byNamedLine (AC);
} minus \polygonEFD =
\drawFromCurrentPicture[middle][polygonABDC]{
draw byNamedPolygon (ABGC, CDG);
}$,\\
and $\polygonAFDC minus \polygonABC =
\drawFromCurrentPicture[middle][polygonEFDC]{
draw byNamedPolygon (EFDG, CDG);
}$;
$\therefore \polygonABDC = \polygonEFDC$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXXVI. Theor.}, reference=prop:I.XXXVI]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, J, I, d[];
numeric h;
h := 3u;
d1 := (3/2u, 0);
d2 := (2/3u, -h);
d3 := (-8/3u, -h);
d4 := (-1/2u, -h);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
E := C shifted -d3;
F := D shifted -d3;
G := E shifted d4;
H := F shifted d4;
I := (B--D) intersectionpoint (C--E);
J := (E--G) intersectionpoint (D--F);
draw byPolygon(A,B,I,C)(byred);
draw byPolygon(C,D,I)(byred);
draw byPolygon(I,D,J,E)(byblue);
draw byPolygon(E,F,J)(byyellow);
draw byPolygon(J,F,H,G)(byyellow);
byLineDefine(C, E, byyellow, 0, 0);
byLineDefine(D, F, black, 1, 0);
byLineDefine(C, D, black, 0, 0);
byLineDefine(E, F, byred, 0, 0);
draw byNamedLineSeq(0)(CE,EF,DF,CD);
draw byLineFull(G, H, byblue, 0, 0)(E, F, 1, 1, 0);
}
\drawCurrentPictureInMargin
\problemNP{P}{arallelograms}{(\drawFromCurrentPicture[bottom][polygonABDC]{
draw byNamedPolygon(ABIC, CDI);
}~and~\drawFromCurrentPicture[bottom][polygonEFHG]{
draw byNamedPolygon(EFJ, JFHG);
}) on equal bases, and between the same parallels are equal.}
\startCenterAlign
Draw \drawUnitLine{CE} and \drawUnitLine{DF}\\
$\drawUnitLine{CD} = \drawUnitLine{GH} = \drawUnitLine{EF}$ by (\inpropL[prop:I.XXXIV], and hyp.);
$\therefore \drawUnitLine{CD} = \mbox{ and } \parallel \drawUnitLine{EF}$;
$\therefore \drawUnitLine{CE} = \mbox{ and } \parallel \drawUnitLine{DF}$ \inprop[prop:I.XXXV]
And therefore
\drawFromCurrentPicture[bottom][polygonCDFE]{
draw byNamedPolygon(CDI, IDJE, EFJ);
}
is a parallelogram:
but $\polygonABDC = \polygonCDFE = \polygonEFHG$ \inprop[prop:I.XXXV]
$\therefore \polygonABDC = \polygonEFHG$ \inax[ax:I].
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXXVII. Theor.}, reference=prop:I.XXXVII]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, I, d[];
d1 := (3/2u, 0);
d2 := (3/4u, -3u);
d3 := (-7/4u, -3u);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
E := C shifted -d3;
F := D shifted -d3;
G := (B--D) intersectionpoint (C--E);
H := 11/10[F, A];
I := 11/10[A, F];
draw byPolygon(A,B,C)(byblue);
draw byPolygon(B,C,G)(byyellow);
draw byPolygon(C,D,G)(byyellow);
draw byPolygon(D,G,E)(black);
draw byPolygon(E,F,D)(byred);
draw byLine(B, D, byred, 0, 0);
draw byLine(E, C, byblue, 0, 0);
byLineDefine(A, C, byred, 1, 0);
byLineDefine(F, D, byblue, 1, 0);
byLineDefine(C, D, black, 0, 0);
draw byNamedLineSeq(0)(AC,CD,FD);
draw byLine(H, I, black, 1, 0);
}
\drawCurrentPictureInMargin
\problemNP{T}{riangles}{
\drawFromCurrentPicture[bottom][polygonBCD]{
draw byNamedPolygon(BCG, CDG);
} and~\drawFromCurrentPicture[bottom][polygonCDE]{
draw byNamedPolygon(DGE, CDG);
}
on the same base (\drawUnitLine{CD}) and between the same parallels are equal.
}
\startCenterAlign
$\left.\eqalign{\mbox{Draw} \drawUnitLine{AC} &\parallel \drawUnitLine{BD}\cr
\drawUnitLine{FD} &\parallel \drawUnitLine{EC}}\right\}\mbox{\inprop[prop:I.XXXI]}$
Produce \drawUnitLine{HI}.
\drawFromCurrentPicture[bottom][polygonABDC]{
draw byNamedPolygon(ABC, BCG, CDG);
}
and
\drawFromCurrentPicture[bottom][polygonEFDC]{
draw byNamedPolygon(DGE, CDG, EFD);
}
are parallelograms on the same base and between the same parallels, and therefore equal. \inprop[prop:I.XXXV]
$\therefore \left\{\eqalign{\polygonABDC &= \mbox{ twice } \polygonBCD\cr\polygonEFDC &= \mbox{ twice } \polygonCDE\cr}\right\}$ \inprop[prop:I.XXXIV]
$\therefore \polygonBCD = \polygonCDE$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXXVIII. Theor.}, reference=prop:I.XXXVIII]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, J, I, d[];
numeric h;
h := 3u;
d1 := (3/2u, 0);
d2 := (2/3u, -h);
d3 := (-8/3u, -h);
d4 := (-1/2u, -h);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
E := C shifted -d3;
F := D shifted -d3;
G := E shifted d4;
H := F shifted d4;
I := (xpart(A), ypart(C));
J := (xpart(F), ypart(C));
draw byPolygon(A,B,C)(byyellow);
draw byPolygon(B,C,D)(byred);
draw byPolygon(E,F,H)(black);
draw byPolygon(E,G,H)(byblue);
draw byLine(B, D, byblue, 0, 0);
draw byLine(E, G, byred, 0, 0);
byLineDefine(A, C, byblue, 1, 0);
byLineDefine(F, H, byred, 1, 0);
byLineDefine(A, F, black, 1, 0);
draw byNamedLineSeq(0)(AC,AF,FH);
draw byLine(I, J, black, 1, 0);
}
\drawCurrentPictureInMargin
\problemNP{T}{riangles}{(\drawFromCurrentPicture[bottom][polygonBCD]{draw byNamedPolygon(BCD);} and \drawFromCurrentPicture[bottom][polygonEGH]{draw byNamedPolygon(EGH);}) on equal bases and between the same parallels are equal.}
\startCenterAlign
$\left.\eqalign{\mbox{Draw} \drawUnitLine{AC} &\parallel \drawUnitLine{BD}\cr
\mbox{and} \drawUnitLine{FH} &\parallel \drawUnitLine{EG}}\right\}\mbox{\inprop[prop:I.XXXI]}$\\
$\drawFromCurrentPicture[bottom][polygonABDC]{
draw byNamedPolygon(ABC, BCD);
} = \drawFromCurrentPicture[bottom][polygonEFHG]{
draw byNamedPolygon(EFH, EGH);
}$ \inprop[prop:I.XXXVI];
but $\polygonABDC = \mbox{ twice } \polygonBCD$ \inprop[prop:I.XXXIV],\\
and $\polygonEFHG = \mbox{ twice } \polygonEGH$ \inprop[prop:I.XXXIV],
$\therefore \polygonBCD = \polygonEGH$ \inax[ax:VII].
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XXXIX. Theor.}, reference=prop:I.XXXIX]
\defineNewPicture{
pair A, B, C, D, E, F, G;
A := (0, 0);
B := A shifted (5/2u, 0);
C := A shifted (u, -5/2u);
D := C shifted (3u, 0);
E = whatever[A, D] = whatever[B, C];
F := 5/4[C, B];
G := 6/4[C, B];
draw byPolygon(A,B,E)(byred);
draw byPolygon(A,E,C)(byyellow);
draw byPolygon(B,E,D)(black);
draw byPolygon(E,C,D)(byyellow);
draw byPolygon(F,B,D)(byblue);
byLineDefine(A, F, byred, 0, 0);
byLineDefine(D, F, byyellow, 0, 0);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(C, D, black, 0, 0);
byLineDefine(C, G, black, 1, 0);
draw byNamedLineSeq(0)(AB,AF,DF,CD,CG);
}
\drawCurrentPictureInMargin
\problemNP{E}{qual}{triangles
\drawFromCurrentPicture[bottom][polygonADC]{
draw byNamedPolygon(AEC, ECD);
} and~\drawFromCurrentPicture[bottom][polygonBDC]{
draw byNamedPolygon(BED, ECD);
}
on the same base (\drawUnitLine{CD}) and on the same side of it, are between the same parallels.}
\startCenterAlign
If \drawUnitLine{AB}, which joins the vertices of the triangles, be not $\parallel \drawUnitLine{CD}$, draw $\drawUnitLine{AF} \parallel \drawUnitLine{CD}$ \inprop[prop:I.XXXI], meeting \drawUnitLine{CG}.
Draw \drawUnitLine{DF}.
Because $\drawUnitLine{AF} \parallel \drawUnitLine{CD}$ (const.)\\
$\polygonADC =
\drawFromCurrentPicture[bottom][polygonFDC]{
draw byNamedPolygon(BED, ECD, FBD);
}$ \inprop[prop:I.XXXVII];\\
but $\polygonADC = \polygonBDC$ (hyp.);
$\therefore \polygonBDC = \polygonFDC$, a part equal to the whole, which is absurd.
$\therefore \drawUnitLine{AF} \nparallel \drawUnitLine{CD}$; and in the same manner it can be demonstrated, that no other line except \drawUnitLine{AB} is $\parallel \drawUnitLine{CD}$; $\therefore \drawUnitLine{AB} \parallel \drawUnitLine{CD}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XL. Theor.}, reference=prop:I.XL]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, d;
A := (0, 0);
B := A shifted (5/2u, 0);
C := A shifted (-u, -9/4u);
d := (2u, 0);
D := C shifted d;
E := B shifted (-1/2u, -9/4u);
F := E shifted d;
G := 5/4[E, B];
H := 2[B, G];
draw byPolygon(A,C,D)(byyellow);
draw byPolygon(B,E,F)(byred);
draw byPolygon(G,B,F)(byblue);
draw byLine(E, H, black, 1, 0);
byLineDefine(A, G, byred, 0, 0);
byLineDefine(F, G, byyellow, 0, 0);
byLineDefine(A, B, byblue, 0, 0);
byLineDefine(C, D, black, 0, 0);
byLineDefine(E, F, black, 0, 0);
byLineDefine(D, E, byblue, 1, 0);
draw byNamedLineSeq(0)(AB,AG,FG,EF,DE,CD);
}
\drawCurrentPictureInMargin
\problemNP{E}{qual}{triangles
(\drawFromCurrentPicture[bottom][polygonACD]{
draw byNamedPolygon (ACD);
draw byNamedLineFull(A, A, 1, 1, 0)(CD);
}
and
\drawFromCurrentPicture[bottom][polygonBEF]{
draw byNamedPolygon (BEF);
draw byNamedLineFull(B, B, 1, 1, 0)(EF);
}) on equal bases, and on the same side, are between the same parallels.}
\startCenterAlign
If \drawSizedLine{AB} which joins the vertices of triangles be not $\parallel \drawSizedLine{CD,DE,EF}$,\\
draw \drawSizedLine{AG} $\parallel \drawSizedLine{CD,DE,EF}$ \inprop[prop:I.XXXI], meeting \drawSizedLine{EH}.\\
Draw \drawSizedLine{FG}.
Because $\drawSizedLine{AG} \parallel \drawSizedLine{CD,DE,EF}$ (const.)\\
$\polygonACD =
\drawFromCurrentPicture[bottom][polygonGEF]{
draw byNamedPolygon (BEF, GBF);
draw byNamedLineFull(G, G, 1, 1, 0)(EF);
}$ but $\polygonACD = \polygonBEF$
$\therefore \polygonBEF = \polygonGEF$, a part equal to the whole, which is absurd.
$\therefore \drawSizedLine{AG} \nparallel \drawSizedLine{CD,DE,EF}$: and in the same manner it can be demonstrated, that no other line except \drawSizedLine{AB} is $\parallel \drawSizedLine{CD,DE,EF}$: $\therefore \drawSizedLine{AB} \parallel \drawSizedLine{CD,DE,EF}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XLI. Theor.}, reference=prop:I.XLI]
\defineNewPicture{
pair A, B, C, D, E, F, G, d;
A := (0, 0);
d := (2u, 0);
B := A shifted d;
C := B shifted (2u, 0);
D := A shifted (4/3u, -3u);
E := D shifted d;
F = whatever[B, E] = whatever[D, C];
G = whatever[A, E] = whatever[D, C];
draw byPolygon(A,B,F,G)(byyellow);
draw byPolygon(G,F,E)(byyellow);
draw byPolygon(A,G,D)(byblue);
draw byPolygon(D,E,G)(byblue);
draw byPolygon(C,F,E)(byred);
draw byLine(A, E, byred, 0, 0);
draw byLineFull(A, C, black, 1, 0)(D, E, 1, 1, 0);
draw byLineFull(D, E, black, 0, 0)(A, C, 1, 1, 0);
}
\drawCurrentPictureInMargin
\problemNP[2]{I}{f}{a parallelogram
\drawFromCurrentPicture[bottom][polygonABED]{
draw byNamedPolygon(ABFG,GFE,AGD,DEG);
}
and a triangle
\drawFromCurrentPicture[bottom][polygonCED]{
draw byNamedPolygon(GFE,DEG,CFE);
}
are upon the same base \drawUnitLine{DE} and between the same parallels \drawUnitLine{AC} and \drawUnitLine{DE}, the parallelogram is double the triangle.}
\startCenterAlign
Draw \drawUnitLine{AE} the diagonal;
Then
$\drawFromCurrentPicture[bottom][polygonAED]{
draw byNamedPolygon(AGD,DEG);
} = \polygonCED$ \inprop[prop:I.XXXVII]\\
$\polygonABED = \mbox{ twice } \polygonAED$ \inprop[prop:I.XXXIV]
$\therefore \polygonABED = \mbox{ twice } \polygonCED$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XLII. Theor.}, reference=prop:I.XLII]
\defineNewPicture{
pair A, B, C, D, E, F, G, d;
A := (0, 0);
d := (2u, 0);
B := A shifted d;
C := B shifted (2u, 0);
D := A shifted (u, -3u);
E := D shifted d;
F := (B--E) intersectionpoint (D--C);
G := 2[D, E];
draw byPolygon(A,B,F,D)(byyellow);
draw byPolygon(D,E,F)(byyellow);
draw byPolygon(C,F,E)(byblue);
draw byPolygon(E,C,G)(black);
draw byAngleWithName(B, E, D, byblue, 0)(E);
draw byAngleWithName(B, E, D, byyellow, 0)(EE) shifted (xpart(G)-xpart(E), -ypart(D)-u);
draw byLine(A, D, byred, 1, 0);
draw byLine(B, E, byred, 0, 0);
draw byLine(C, E, byyellow, 0, 0);
draw byLine(A, C, byblue, 0, 0);
draw byLine(D, E, black, 0, 0);
draw byLine(E, G, black, 1, 0);
}
\drawCurrentPicture
\problemNP[2]{T}{o}{construct a parallelogram equal to a given triangle
\drawPolygon[bottom][polygonCDG]{DEF,CFE,ECG} and having an angle equal to a given rectilinear angle \drawAngle{EE}.}
\startCenterAlign
Make $\drawUnitLine{DE} = \drawUnitLine{EG}$ \inprop[prop:I.X]\\
Draw \drawUnitLine{CE}.
Make $\drawAngle{E} = \drawAngle{EE}$ \inprop[prop:I.XXIII]\\
Draw $\left\{\eqalign{\drawUnitLine{AD} &\parallel \drawUnitLine{BE}\cr\drawUnitLine{AC} &\parallel \drawUnitLine{DE}\cr}\right\}$ \inprop[prop:I.XXXI]
$\drawFromCurrentPicture[bottom][polygonABED]{
draw byNamedPolygon(ABFD,DEF);
}
= \mbox{ twice }
\drawFromCurrentPicture[bottom][polygonCED]{
draw byNamedPolygon(DEF,CFE);
}$ \inprop[prop:I.XLI]\\
but $\polygonCED =
\drawFromCurrentPicture[bottom][polygonDCG]{
draw byNamedPolygon(ECG);
}$ \inprop[prop:I.XXXVIII]
$\therefore \polygonABED = \polygonCDG$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XLIII. Theor.}, reference=prop:I.XLIII]
\defineNewPicture[1/2]{
pair A, B, C, D, E, F, G, H, I, d[];
path q[];
d1 := (5/2u, 0);
d2 := (-u, -3u);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
E := 2/5[A, D];
q1 := (E shifted d1) -- (E shifted -d1);
q2 := (E shifted d2) -- (E shifted -d2);
F := q1 intersectionpoint (A--C);
G := q1 intersectionpoint (B--D);
H := q2 intersectionpoint (A--B);
I := q2 intersectionpoint (C--D);
draw byPolygon(A,E,H)(byyellow);
draw byPolygon(A,E,F)(byyellow);
draw byPolygon(H,B,G,E)(byblue);
draw byPolygon(F,C,I,E)(black);
draw byPolygon(I,D,E)(byred);
draw byPolygon(G,D,E)(byred);
}
\drawCurrentPictureInMargin
\problemNP{T}{he}{complements
\drawFromCurrentPicture[bottom][polygonHBGE]{
draw byNamedPolygon(HBGE);
}
and
\drawFromCurrentPicture[bottom][polygonFCIE]{
draw byNamedPolygon(FCIE);
}
of the parallelograms which are about the diagonal of a parallelogram are equal.}
\startCenterAlign
$\drawFromCurrentPicture[bottom][polygonADC]{
draw byNamedPolygon(AEF,FCIE,IDE);
} = \drawFromCurrentPicture[bottom][polygonABD]{
draw byNamedPolygon(AEH,HBGE,GDE);
}$ \inprop[prop:I.XXXIV]\\
and $\drawFromCurrentPicture[bottom][polygonAEFpIDE]{
draw byNamedPolygon(AEF,IDE);
} = \drawFromCurrentPicture[bottom][polygonAEHpGDE]{
draw byNamedPolygon(AEH,GDE);
}$ \inprop[prop:I.XXXIV]
$\therefore \polygonFCIE = \polygonHBGE$ \inax[ax:III]
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XLIV. Prob.}, reference=prop:I.XLIV]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, I, J, K, L, d[];
path q[];
d1 := (3u, 0);
d2 := (-3/2u, -3u);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
E := 2/5[C, B];
q1 := (E shifted d1) -- (E shifted -d1);
q2 := (E shifted d2) -- (E shifted -d2);
F := q1 intersectionpoint (A--C);
G := q1 intersectionpoint (B--D);
H := q2 intersectionpoint (A--B);
I := q2 intersectionpoint (C--D);
J := A shifted (-1/2u, 0);
K := J shifted (2(xpart(A)-xpart(H)), 0);
L := (xpart(1/3[J, K]), ypart(F));
draw byPolygon(J,K,L)(byred);
draw byPolygon(A,H,E,F)(byyellow);
draw byPolygon(E,G,D,I)(byblue);
draw byAngleWithName(A, F, E, byblue, 0)(F);
draw byAngleWithName(F, E, I, byred, 0)(E);
draw byAngleWithName(E, I, D, black, 0)(I);
draw byAngleWithName(A, F, E, byyellow, 0)(FF) shifted d1 shifted (0, ypart(D)-ypart(F));
draw byLine(B, E, byred, 0, 0);
draw byLine(E, C, black, 0, 1);
byLineDefine(A, F, byred, 1, 0);
byLineDefine(F, C, black, 0, 1);
draw byLine(H, E, byblue, 1, 0);
byLineDefine(B, G, byyellow, 0, 0);
byLineDefine(A, H, byblue, 0, 0);
byLineDefine(H, B, black, 0, 0);
byLineDefine(F, E, black, 1, 0);
byLineDefine(E, G, black, 0, 0);
byLineDefine(C, D, byyellow, 1, 0);
draw byNamedLineSeq(0)(FE,EG,BG,HB,AH,AF,FC,CD);
}
\drawCurrentPicture
\initialIndentation{8}
\problem{T}{o}{a given straight line (\drawUnitLine{EG}) to apply a parallelogram equal to a given triangle (\drawFromCurrentPicture[bottom][polygonJKL]{draw byNamedPolygon(JKL);}), and having an angle equal to a given rectilinear angle (\drawAngle{FF}).}
\startCenterAlign
Make $\drawFromCurrentPicture[bottom][polygonAHEF]{draw byNamedPolygon(AHEF);} = \polygonJKL$ with $\drawAngle{F} = \drawAngle{FF}$ \inprop[prop:I.XLII]\\
and having one of its sides \drawUnitLine{FE} conterminous with and in continuation of \drawUnitLine{EG}.
Produce \drawUnitLine{AH} till it meets $\drawUnitLine{BG} \parallel \drawUnitLine{HE}$ draw \drawUnitLine{BE} produce it till it meets \drawUnitLine{AF} continued; draw $\drawUnitLine{CD} \parallel \drawUnitLine{FE,EG}$ meeting \drawUnitLine{BG} produced and produce \drawUnitLine{HE}.
$\polygonAHEF = \drawFromCurrentPicture[bottom][polygonEGDI]{draw byNamedPolygon(EGDI);}$ \inprop[prop:I.XLIII]\\
but $\polygonAHEF = \polygonJKL$ (const.)
$\therefore \polygonEGDI = \polygonAHEF$;
and $\drawAngle{F} = \drawAngle{E} =\drawAngle{I} = \drawAngle{FF}$ (\inpropL[prop:I.XXIX] and const.)
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XLV. Prob.}, reference=prop:I.XLV]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, I, J, K, L, M, d[];
numeric a, h[], b[], s[];
a := 15;
A := (0, 0);
B := A shifted (0, 2u);
C := A shifted (4/3u, u);
D := A shifted (2u, -3/2u);
E := A shifted (-6/5u, -u);
b1 := arclength(B--C);
h1 := distanceToLine(A, B--C);
s1 := (b1 * h1)/2;
b2 := arclength(C--D);
h2 := distanceToLine(A, C--D);
s2 := (b2 * h2)/2;
b3 := arclength(D--E);
h3 := distanceToLine(A, D--E);
s3 := (b3 * h3)/2;
d1 := (0, ypart(D)-u);
d2 := (0, -b3/2) rotated -a;
d3 := (h3*(1/cosd(a)), 0);
F := (-u, 0) shifted d1;
G := F shifted d3;
H := F shifted d2;
I := G shifted d2;
d4 := (2*(s2/b3)*(1/cosd(a)), 0);
J := G shifted d4;
K := J shifted d2;
d5 := (2*(s1/b3)*(1/cosd(a)), 0);
L := J shifted d5;
M := L shifted d2;
draw byPolygon(A,B,C)(byred);
draw byPolygon(A,C,D)(byyellow);
draw byPolygon(A,D,E)(byblue);
byLineDefine(A, C, byblue, 0, 0);
byLineDefine(A, D, byred, 0, 0);
draw byNamedLineSeq(0)(AC,AD);
draw byPolygon(F,G,I,H)(byblue);
draw byPolygon(G,J,K,I)(byyellow);
draw byPolygon(J,L,M,K)(byred);
draw byAngleWithName(G, I, H, byyellow, 0)(I);
draw byAngleWithName(J, K, I, black, 0)(K);
draw byAngleWithName(L, M, K, byblue, 0)(M);
draw byAngleWithName(L, M, K, byred, 0)(O) shifted (u, 0);
draw byLine(G, I, byred, 0, 0);
draw byLine(J, K, byblue, 0, 0);
}
\drawCurrentPictureInMargin
\problemNP[2]{T}{o}{construct a parallelogram equal to a given rectilinear figure (\drawPolygon[middle][polygonABCDE]{ABC,ACD,ADE}) on an angle, equal to a given rectilinear angle (\drawAngle{O}).}
\startCenterAlign
Draw \drawUnitLine{AD} and \drawUnitLine{AC} dividing the rectilinear figure into triangles.
Construct $\drawPolygon{FGIH} = \drawPolygon{ADE}$\\
having $\drawAngle{I} = \drawAngle{O}$ \inprop[prop:I.XLII]
to \drawUnitLine{GI} apply $\drawPolygon{GJKI} = \drawPolygon{ACD}$\\
having $\drawAngle{K} = \drawAngle{O}$ \inprop[prop:I.XLIV]
to \drawUnitLine{JK} apply $\drawPolygon{JLMK} = \drawPolygon{ABC}$\\
having $\drawAngle{M} = \drawAngle{O}$ \inprop[prop:I.XLIV]
$\therefore \drawPolygon[middle][polygonFLMH]{FGIH,GJKI,JLMK} = \polygonABCDE$
and \polygonFLMH is a parallelogram. (\inpropL[prop:I.XXIX], \inpropL[prop:I.XIV], \inpropL[prop:I.XXX])\\
having $\drawAngle{M} = \drawAngle{O}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XLVI. Prob.}, reference=prop:I.XLVI]
\defineNewPicture{
pair A, B, C, D;
numeric d;
d := 7/2u;
A := (0, 0);
B := A shifted (d, 0);
C := A shifted (0, -d);
D := A shifted (d, -d);
draw byAngleWithName(B, A, C, black, 0)(A);
draw byAngleWithName(D, B, A, byblue, 0)(B);
draw byAngleWithName(C, D, B, byred, 0)(D);
draw byAngleWithName(A, C, D, byyellow, 0)(C);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(B, D, byyellow, 0, 0);
byLineDefine(D, C, black, 0, 0);
byLineDefine(C, A, byblue, 0, 0)
draw byNamedLineSeq(0)(AB,BD,DC,CA);
}
\drawCurrentPictureInMargin
\problemNP{U}{pon}{a given straight line (\drawUnitLine{DC}) to construct a square.}
\startCenterAlign
Draw $\drawUnitLine{CA} \perp \mbox{ and } = \drawUnitLine{DC}$ (\inpropL[prop:I.XI], \inpropL[prop:I.III])
Draw $\drawUnitLine{AB} \parallel \drawUnitLine{DC}$, and meeting \drawUnitLine{BD} drawn $\parallel \drawUnitLine{CA}$.
In
\drawFromCurrentPicture[bottom][polygonABDC]{
startTempAngleScale(angleScale*3/4);
draw byNamedAngle(A,B,C,D);
draw byNamedLineSeq(0)(AB,BD,DC,CA);
stopTempAngleScale;
}
$\drawUnitLine{CA} = \drawUnitLine{DC}$ (const.)\\
$\drawAngle{C} = \mbox{right angle}$ (const.)
$\therefore \drawAngle{D} = \drawAngle{C} = \mbox{a right angle}$ \inprop[prop:I.XXIX], and the remaining sides and angles must be equal, \inprop[prop:I.XXXIV]
And $\therefore \polygonABDC$ is a square. \indef[def:XXVII]
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XLVII. Theor.}, reference=prop:I.XLVII]
\defineNewPicture[1/2]{
pair A, B, C, D, E, F, G, H, I, J, K, L, M, d[];
A := (0, 0);
B := A shifted (-7/10u, -8/7u);
C = whatever[A, A shifted ((A-B) rotated 90)] = whatever[B, B shifted dir(0)];
d1 := (B-A) rotated -90;
D := A shifted d1;
E := B shifted d1;
d2 := (A-C) rotated -90;
F := C shifted d2;
G := A shifted d2;
d3 := (C-B) rotated -90;
H := B shifted d3;
I := C shifted d3;
J = whatever[A, A shifted dir(90)];
J = whatever[B, C];
K = whatever[A, A shifted dir(90)];
K = whatever[H, I];
L = whatever[B, F];
L = whatever[A, C];
M = whatever[A, I];
M = whatever[B, C];
draw byPolygon(A,B,E,D)(black);
draw byPolygon(L,A,G,F)(byred);
draw byPolygon(C,L,F)(byred);
draw byPolygon(J,M,I,K)(byblue);
draw byPolygon (M,C,I)(byblue);
draw byPolygon(B,J,K,H)(byyellow);
draw byAngle(F, C, A, byyellow, 0);
draw byAngle(B, C, I, byyellow, 0);
draw byAngle(A, C, B, black, 0);
draw byLine(A, K, black, 1, 0);
draw byLineFull(B, F, black, 0, 0)(C, G, 1, 1, 0);
draw byLineFull(A, I, black, 0, 0)(C, H, 1, 1, 0);
draw byLineFull(C, F, byblue, 1, 0)(C, F, 0, 0, 1/2);
draw byLineFull(C, I, byred, 1, 0)(C, I, 0, 0, -1/2);
byLineDefine(A, B, byyellow, 0, 0);
byLineDefine(B, C, byred, 0, 0);
byLineDefine(C, A, byblue, 0, 0);
draw byNamedLineSeq(-1/2)(AB,BC,CA);
byLineDefineWithName (C, A, black, 0, 0)(CAb);
byLineStylize (M, M, 1, 0, -1) (CAb);
byLineDefineWithName (A, M, black, 0, 0)(AMb);
byLineStylize (C, C, 0, 1, -1) (AMb);
byLineDefineWithName (B, C, black, 0, 0)(BCb);
byLineStylize (L, L, 0, 1, -1) (BCb);
byLineDefineWithName (L, B, black, 0, 0)(BLb);
byLineStylize (C, C, 1, 0, -1) (BLb);
}
\drawCurrentPictureInMargin
\problemNP{I}{n}{a right angled triangle \drawLine[bottom][triangleABC]{CA,BC,AB} the square on the hypotenuse \drawUnitLine{BC} is equal to the sum of the squares of the sides (\drawUnitLine{CA} and \drawUnitLine{AB}).}
\startCenterAlign
On \drawUnitLine{BC}, \drawUnitLine{CA}, \drawUnitLine{AB} describe squares, \inprop[prop:I.XLVI]
Draw $\drawUnitLine{AK} \parallel \drawUnitLine{CI}$ \inprop[prop:I.XXXI]\\
also draw \drawUnitLine{BF} and \drawUnitLine{AI}.\\
$\drawAngle{BCI} = \drawAngle{FCA}$,
To each add \drawAngle{ACB} $\therefore \drawAngle{BCI,ACB} = \drawAngle{FCA,ACB}$,\\
$\drawUnitLine{BC} = \drawUnitLine{CI}$ and $\drawUnitLine{CA} = \drawUnitLine{CF}$;
$\therefore
\drawFromCurrentPicture[middle][polygonAFC]{
draw byNamedPolygon(MCI);
draw byNamedAngle(ACB);
draw byNamedLine(CAb,AMb);
}
=
\drawFromCurrentPicture[middle][polygonBLC]{
draw byNamedPolygon(CLF);
draw byNamedAngle(ACB);
draw byNamedLine(BCb,BLb);
}
$.
Again, because $\drawUnitLine{AB} \parallel \drawUnitLine{CF}$\\
$\drawFromCurrentPicture[middle][polygonACFG]{draw byNamedPolygon(LAGF,CLF);} = \mbox{ twice } \polygonBLC$,\\
and $\drawFromCurrentPicture[middle][polygonJMCK]{draw byNamedPolygon(JMIK,MCI);} = \mbox{ twice } \polygonAFC$;
$\therefore \polygonACFG = \polygonJMCK$.
In the same manner it can be shown that $\drawFromCurrentPicture[middle][polygonABED]{draw byNamedPolygon(ABED);} = \drawFromCurrentPicture[middle][polygonBJKH]{draw byNamedPolygon(BJKH);}$;
hence $\drawFromCurrentPicture[middle][polygonABEDpACFG]{draw byNamedPolygon(ABED,LAGF,CLF);} = \drawFromCurrentPicture[middle][polygonBCIH]{draw byNamedPolygon(JMIK,MCI,BJKH);}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop XLVIII. Theor.}, reference=prop:I.XLVIII]
\defineNewPicture{
pair A, B, C, D;
numeric d;
d := 7/4u;
A := (0, 0);
B := A shifted (0, 3u);
C := A shifted (d, 0);
D := A shifted (-d, 0);
draw byAngle(B, A, C, byred, 0);
draw byAngle(D, A, B, byyellow, 0);
draw byLine(A, B, byblue, 0, 0);
byLineDefine(A, C, black, 0, 0);
byLineDefine(A, D, black, 1, 0);
byLineDefine(B, C, byred, 0, 0);
byLineDefine(B, D, byred, 1, 0);
draw byNamedLineSeq(0)(AC,AD,BD,BC);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{the square of one side (\drawUnitLine{BC}) of a triangle is equal to the squares of the other two sides (\drawUnitLine{AB} and \drawUnitLine{AC}), the angle (\drawAngle{DAB}) subtended by that side is a right angle.}
\startCenterAlign
Draw $\drawUnitLine{AD} \perp \drawUnitLine{AB}$ and $= \drawUnitLine{AC}$ (\inpropL[prop:I.XI], \inpropL[prop:I.III])\\
and also draw \drawUnitLine{BD} also.
Since $\drawUnitLine{AD} = \drawUnitLine{AC}$ (const.)\\
$\drawUnitLine{AD}^2 = \drawUnitLine{AC}^2$;
$\therefore \drawUnitLine{AD}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{AC}^2 + \drawUnitLine{AB}^2$
but $\drawUnitLine{AD}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BD}^2$ \inprop[prop:I.XLVII],\\
and $\drawUnitLine{AC}^2 + \drawUnitLine{AB}^2 = \drawUnitLine{BC}^2$ (hyp.)
$\therefore \drawUnitLine{BD}^2 = \drawUnitLine{BC}^2$,
$\therefore \drawUnitLine{BD} = \drawUnitLine{BC}$;
and $\therefore \drawAngle{DAB} = \drawAngle{BAC}$ \inprop[prop:I.VIII],
consequently \drawAngle{BAC} is a right angle.
\stopCenterAlign
\qed
\stopProposition
\stopbook
\startbook[title={Book 2}]
\startDefinition[title={Definition I},reference=def:II.I]
\defineNewPicture{
pair A, B, C, D;
numeric w, h;
w := 7/2u;
h := 3u;
A := (0, 0);
B := (w, 0);
C := (0, h);
D := (w, h);
draw byPolygon(A,B,D,C)(byblue);
byLineDefine(A, B, black, 0, 0);
byLineDefine(A, C, byred, 0, 0);
draw byNamedLineSeq(0)(AB,AC);
}
\drawCurrentPictureInMargin
\problemNP{A}{rectangle}{or a right angled parallelogram is said to be contained by any two of its adjacent or conterminous sides.}
Thus: the right angled parallelogram \drawFromCurrentPicture{draw byNamedPolygon (ABDC);} is said to be contained by the sides \drawUnitLine{AB} and \drawUnitLine{AC}; or it may be briefly designated by $\drawUnitLine{AB} \cdot \drawUnitLine{AC}$.
If the adjacent sides are equal; i. e. $\drawUnitLine{AB} = \drawUnitLine{AC}$, then $\drawUnitLine{AB} \cdot \drawUnitLine{AC}$ which is the expression for the rectangle undet \drawUnitLine{AB} and \drawUnitLine{AC} is square, and
is equal to $\left\{\eqalign{
\drawUnitLine{AB} \cdot \drawUnitLine{AC}&\mbox{ or } \drawUnitLine{AC}^2\cr
\drawUnitLine{AB} \cdot \drawUnitLine{AC}&\mbox{ or } \drawUnitLine{AC}^2}\right.$
\stopDefinition
\vfill\pagebreak
\startDefinition[title={Definition II},reference=def:II.II]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, I, d[];
d1 := (3u, 0);
d2 := (1/2u, 3u);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := A shifted d1 shifted d2;
E := 2/3[A, B];
F := 2/3[C, D];
G := 2/3[A, C];
H := 2/3[B, D];
I = whatever[E, F] = whatever[G, H];
draw byPolygon(A,E,I,G)(byblue);
draw byPolygon(E,B,H,I)(byyellow);
draw byPolygon(G,I,F,C)(byyellow);
draw byPolygon(I,H,D,F)(byred);
}
\drawCurrentPictureInMargin
\problemNP[4]{I}{n}{a parallelogram, the figure composed of one of the parallelograms about the diagonal together with the two complements, is called a Gnomon.}
Thus \drawFromCurrentPicture[middle][gnomonA]{draw byNamedPolygon (AEIG);draw byNamedPolygon (EBHI);draw byNamedPolygon (GIFC);} and \drawFromCurrentPicture[middle][gnomonB]{draw byNamedPolygon (EBHI);draw byNamedPolygon (GIFC);draw byNamedPolygon (IHDF);} are called Gnomons.
\stopDefinition
\vfill\pagebreak
\startProposition[title={Prop. I. prob.},reference=prop:II.I]
\defineNewPicture{
pair B,C,D,E,G,H,K,L;
numeric w, h;
w := 7/2u;
h := 3u;
G := (0, 0);
H := (w, 0);
B := (0, h);
C := (w, h);
K := 2/5[G, H];
D := 2/5[B, C];
L := 3/4[G, H];
E := 3/4[B, C];
draw byPolygon(G,K,D,B)(byyellow);
draw byPolygon(K,L,E,D)(byblue);
draw byPolygon(L,H,C,E)(byred);
draw byLine(K, D, black, 1, 0);
draw byLine(L, E, black, 1, 0);
byLineDefine(G, B, black, 0, 0);
byLineDefine(H, C, black, 1, 0);
byLineDefine(G, K, byblue, 0, 0);
byLineDefine(K, L, byred, 0, 0);
byLineDefine(L, H, byyellow, 0, 0);
byLineDefine(B, D, byblue, 1, 0);
byLineDefine(D, E, byred, 1, 0);
byLineDefine(E, C, byyellow, 1, 0);
draw byNamedLineSeq(0)(GK,KL,LH,HC,EC,DE,BD,GB);
}
\drawCurrentPictureInMargin
\problemNP[2]{T}{he}{rectangle contained by two straight lines, one of which is divided into any number of parts,
$\drawProportionalLine{GK,KL,LH} \cdot \drawProportionalLine{GB} = \left\{\eqalign{
&\drawProportionalLine{GB} \cdot \drawProportionalLine{GK}\cr
+&\drawProportionalLine{GB} \cdot \drawProportionalLine{KL}\cr
+&\drawProportionalLine{GB} \cdot \drawProportionalLine{LH}}\right.$\\
is equal to the sum of the rectangles contained by the undivided line, and several parts of the divided line.}
\startCenterAlign
Draw $\drawProportionalLine{GB} \perp \drawProportionalLine{GK,KL,LH} \mbox{ and} = \drawProportionalLine{GB}$ (\inpropL[prop:I.II], \inpropL[prop:I.III]); complete the parallelograms, that is to say,
Draw $\left\{\eqalign{
\drawProportionalLine{BD,DE,EC} &\parallel \drawProportionalLine{GK,KL,LH} \cr
\vcenter{
\nointerlineskip\hbox{\drawProportionalLine{KD}}
\nointerlineskip\hbox{\drawProportionalLine{LE}}
\nointerlineskip\hbox{\drawProportionalLine{HC}}} &\parallel \drawProportionalLine{GB}
}\right\}$ \inprop[prop:I.XXXI]
$\drawPolygon[bottom]{GKDB,KLED,LHCE} =
\drawPolygon[bottom]{GKDB} +
\drawPolygon[bottom]{KLED} +
\drawPolygon[bottom]{LHCE}$
$\drawPolygon[bottom]{GKDB,KLED,LHCE} = \drawProportionalLine{GK,KL,LH} \cdot \drawProportionalLine{GB}$
$\polygonGKDB = \drawProportionalLine{GK} \cdot \drawProportionalLine{GB}$,
$\polygonKLED = \drawProportionalLine{KL} \cdot \drawProportionalLine{GB}$,
$\polygonLHCE = \drawProportionalLine{LH} \cdot \drawProportionalLine{GB}$
$\therefore \drawProportionalLine{GK,KL,LH} \cdot \drawProportionalLine{GB} = \drawProportionalLine{GK} \cdot \drawProportionalLine{GB} + \drawProportionalLine{KL} \cdot \drawProportionalLine{GB} + \drawProportionalLine{LH} \cdot \drawProportionalLine{GB}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. II. theor.},reference=prop:II.II]
\defineNewPicture{
pair A, B, C, D, E, F;
numeric w;
w := 7/2u;
A := (0, w);
B := (w, w);
C := 2/3[A, B];
D := (0, 0);
E := (w, 0);
F := 2/3[D, E];
draw byPolygon(A,C,F,D)(byred);
draw byPolygon(C,B,E,F)(byyellow);
draw byLine(C, F, black, 0, 0);
byLineDefine(A, D, black, 1, 0);
byLineDefine(B, E, black, 1, 0);
byLineDefine(A, C, byblue, 0, 0);
byLineDefine(C, B, byred, 0, 0);
draw byNamedLineSeq(0)(AD,AC,CB,BE);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a straight line be divided into any two parts \drawProportionalLine{AC,CB}, the square of the whole line is equal to the sum of the rectangles contained by the whole line and each of its parts.\\
$\drawProportionalLine{AC,CB}^2 = \left\{\eqalign{
& \drawProportionalLine{AC,CB} \cdot \drawProportionalLine{AC} \cr
+ & \drawProportionalLine{AC,CB} \cdot \drawProportionalLine{CB}
}\right.$
}
\startCenterAlign
Describe \drawPolygon[bottom][polygonABED]{ACFD,CBEF} \inprop[prop:I.XLVI]
Draw \drawProportionalLine{CF} parallel to \drawProportionalLine{AD} \inprop[prop:I.XXXI]
$\polygonABED = \drawProportionalLine{AC,CB}^2$
$\drawPolygon[bottom]{ACFD} = \drawProportionalLine{CF} \cdot \drawProportionalLine{AC} = \drawProportionalLine{AC,CB} \cdot \drawProportionalLine{AC}$
$\drawPolygon[bottom]{CBEF} = \drawProportionalLine{CF} \cdot \drawProportionalLine{CB} = \drawProportionalLine{AC,CB} \cdot \drawProportionalLine{CB}$
$\polygonABED = \drawPolygon[bottom]{ACFD} + \drawPolygon[bottom]{CBEF}$
$\therefore \drawProportionalLine{AC,CB}^2 = \drawProportionalLine{AC,CB} \cdot \drawProportionalLine{AC} + \drawProportionalLine{AC,CB} \cdot \drawProportionalLine{CB}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. III. theor.},reference=prop:II.III]
\defineNewPicture{
pair A, B, C, D, E, F;
numeric w, h;
w := -4u;
h := 11/4u;
A := (0, h);
B := (w, h);
C := (w+h, h);
D := (w+h, 0);
E := (w, 0);
F := (0, 0);
draw byPolygon(A,C,D,F)(byyellow);
draw byPolygon(C,B,E,D)(byred);
byLineDefine(D, F, byred, 0, 0);
byLineDefine(B, C, byblue, 0, 0);
byLineDefine(C, D, byblue, 0, 0);
byLineDefine(D, E, byblue, 0, 0);
byLineDefine(E, B, byblue, 0, 0);
draw byNamedLineSeq(0)(CD,noLine,DF,DE,EB,BC);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a straight line be divided into any two parts \drawProportionalLine{DE,DF}, the rectangle contained by the whole line and either of its parts, is equal to the square of that part, together with the rectangle under the parts.\\
$\drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DE} = \drawProportionalLine{DE}^2 + \drawProportionalLine{DE} \cdot \drawProportionalLine{DF}$, or \\
$\drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DF} = \drawProportionalLine{DF}^2 + \drawProportionalLine{DE} \cdot \drawProportionalLine{DF}$.}
\startCenterAlign
Describe \drawPolygon[bottom]{CBED} \inprop[prop:I.XLVI]
Describe \drawPolygon[bottom]{ACDF} \inprop[prop:I.XXXI]
Then $\drawPolygon[bottom][polygonABEF]{ACDF,CBED} = \polygonCBED + \polygonACFD$, but\\
$\polygonABEF = \drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DE}$ and\\
$\polygonCBED = \drawProportionalLine{DE}^2$, $\polygonACFD = \drawProportionalLine{DE} \cdot \drawProportionalLine{DF}$,
$\therefore \drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DE} = \drawProportionalLine{DE}^2 + \drawProportionalLine{DE} \cdot \drawProportionalLine{DF}$:
In the similar manner it may be readily shown that $\drawProportionalLine{DE,DF} \cdot \drawProportionalLine{DF} = \drawProportionalLine{DF}^2 + \drawProportionalLine{DE} \cdot \drawProportionalLine{DF}$.
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. IV. theor.},reference=prop:II.IV]
\defineNewPicture[1/4]{
pair A, B, C, D, E, F, G, H, K;
numeric w;
w := 9/2u;
A := (0, w);
B := (w, w);
C :=2/3[A, B];
D := (0, 0);
E := (w, 0);
F := 2/3[D, E];
H := 2/3[D, A];
K := 2/3[E, B];
G = whatever[H, K] = whatever[F, C];
draw byPolygon(A,C,G,H)(byyellow);
draw byPolygon(G,K,E,F)(byyellow);
draw byPolygon(D,H,G)(byblue);
draw byPolygon(G,B,C)(byred);
draw byAngle(F, D, G, byyellow, 0);
draw byAngle(F, G, D, byred, 0);
draw byAngle(K, G, B, black, 0);
draw byAngle(G, B, K, byblue, 0);
draw byLine(H, G, byred, 1, 0);
draw byLine(G, K, byred, 0, 0);
draw byLine(C, G, byblue, 1, 0);
draw byLine(F, G, byblue, 0, 0);
byLineDefine(E, K, byblue, 0, 0);
byLineDefine(F, E, byred, 0, 0);
byLineDefine(D, F, byblue, 0, 0);
byLineDefine(K, B, byred, 0, 0);
byLineDefine(G, D, black, 0, 0);
byLineDefine(B, G, black, 1, 0);
draw byNamedLineSeq(-1)(BG,GD,DF,FE,EK,KB);
byLineDefine(A, D, black, 0, 0);
byLineStylize(B, E, 0, 0, -1)(AD);
byLineDefine(B, A, black, 0, 0);
byLineStylize(E, D, 0, 0, -1)(BA);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a straight line be divided into any two parts \drawProportionalLine{DF,FE}, the square of the whole line is equal to the squares of the parts, together with twice the rectangle contained by the parts.\\
$\drawProportionalLine{DF,FE}^2 = \drawProportionalLine{DF}^2 + \drawProportionalLine{FE}^2 + \mbox{twice} \drawProportionalLine{DF} \cdot \drawProportionalLine{FE}$
}
\startCenterAlign
Describe \drawLine[bottom][squareABED]{AD,BA,KB,EK,FE,DF} \inprop[prop:I.XLVI]
draw \drawProportionalLine{BG,GD} \inpost[post:I]\\
and $\left\{\eqalign{
\drawProportionalLine{FG,CG} &\parallel \drawProportionalLine{EK,KB} \cr
\drawProportionalLine{HG,GK} &\parallel \drawProportionalLine{DF,FE}
}\right\}$ \inprop[prop:I.XXXI]\\
$\drawAngle{GBK} = \drawAngle{FDG}$ \inprop[prop:I.V],\\
$\drawAngle{GBK} = \drawAngle{FGD}$ \inprop[prop:I.XXIX],
$\therefore \drawAngle{FDG} = \drawAngle{FGD}$
$\therefore$ by (\inpropL[prop:I.VI], \inpropL[prop:I.XXIX], \inpropL[prop:I.XXXIV])\\ $\drawFromCurrentPicture[bottom][squareFGHD]{
draw byNamedPolygon(DHG);
draw byNamedLineFull(G, G, 1, 0, -1)(DF);
draw byNamedLineFull(D, D, 0, 1, -1)(FG);
} \mbox{ is a square } = \drawProportionalLine{DF}^2$.
For the same reasons \drawFromCurrentPicture[bottom][squareKBCG]{
draw byNamedPolygon(GBC);
draw byNamedLineFull(B, B, 1, 0, -1)(GK);
draw byNamedLineFull(G, G, 0, 1, -1)(KB);
} is a square $= \drawProportionalLine{GK}^2$,
$\drawPolygon[bottom]{ACGH} = \drawPolygon[middle]{GKEF} = \drawProportionalLine{DF} \cdot \drawProportionalLine{GK}$ \inprop[prop:I.XLIII]
But $\drawFromCurrentPicture[bottom][squareABEDf]{
draw byNamedPolygon(GBC,DHG,ACGH,GKEF);
draw byNamedLineFull(G, G, 1, 0, -1)(DF);
draw byNamedLineFull(D, D, 0, 1, -1)(FG);
draw byNamedLineFull(B, B, 1, 0, -1)(GK);
draw byNamedLineFull(G, G, 0, 1, -1)(KB);
} = \squareFGHD + \drawPolygon{ACGH} + \drawPolygon{GKEF} + \squareKBCG$,
$\therefore \drawProportionalLine{DF,FE}^2 = \drawProportionalLine{DF}^2 + \drawProportionalLine{FE}^2 + \mbox{ twice } \drawProportionalLine{DF} \cdot \drawProportionalLine{FE}$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. V. theor.},reference=prop:II.V]
\defineNewPicture[1/4]{
pair A, B, C, D, E, F, G, H, K, L, M;
numeric h;
h := 6u;
A := (0, -h);
B := (0, 0);
C := 1/2[A, B];
D := 2/5[B, C];
E := (1/2h, -1/2h);
F := (1/2h, 0);
G := (xpart(F), ypart(D));
H = whatever[D, G] = whatever[B, E];
K := (xpart(H), ypart(A));
L := (xpart(H), ypart(C));
M := (xpart(H), ypart(B));
draw byPolygon(B,D,H,M)(byblue);
draw byPolygon(D,C,L,H)(byyellow);
draw byPolygon(C,L,K,A)(black);
draw byPolygon(M,H,G,F)(byyellow);
draw byPolygon(H,L,E,G)(byred);
draw byLine(H, G, black, 1, 0);
draw byLine(D, H, byred, 0, 0);
byLineDefine(L, M, black, 1, 0);
byLineDefine(C, L, byred, 0, 0);
byLineDefine(B, D, byred, 0, 0);
byLineDefine(D, C, byblue, 0, 0);
byLineDefine(C, A, byyellow, 0, 0);
byLineDefine(E, B, black, 0, 0);
byLineDefine(A, K, byred, 1, 0);
byLineDefine(K, L, byyellow, 0, 0);
byLineDefine(L, E, byblue, 1, 0);
draw byNamedLineSeq(-1)(BD,DC,CA,AK,KL,LM,CL,LE,EB);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a straight line be divided \drawProportionalLine{BD,DC,CA} into two equal parts and also \drawProportionalLine{BD,DC,CA} into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half that line \\
$\drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA} + \drawProportionalLine{DC}^2 = \drawProportionalLine{CA}^2 = \drawProportionalLine{BD,DC}^2$.
}
\startCenterAlign
Describe \drawPolygon[bottom][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \inprop[prop:I.XLVI],
draw \drawProportionalLine{EB}\\
and $\left\{\eqalign{
\drawProportionalLine{HG,DH} & \parallel \drawProportionalLine{CL,LE} \cr
\drawProportionalLine{LM,KL} & \parallel \drawProportionalLine{BD,DC,CA} \cr
\drawProportionalLine{AK} & \parallel \drawProportionalLine{CL,LE}
}\right\}$ \inprop[prop:I.XXXI]
$\drawPolygon[bottom]{CLKA} = \drawPolygon[bottom]{DCLH,BDHM}$ \inprop[prop:I.XXXVI]
$\drawPolygon[bottom]{MHGF} = \drawPolygon[bottom]{DCLH}$ \inprop[prop:I.XLIII]
$\therefore \mbox{\inax[ax:II] } \drawPolygon[bottom]{DCLH,BDHM,MHGF} = \drawFromCurrentPicture[bottom]{draw byNamedPolygon(DCLH,CLKA) rotated 90;} = \drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA}$
but $\drawPolygon[bottom]{HLEG} = \drawProportionalLine{DC}^2$ \inprop[prop:II.IV]\\
and $\squareCBFE = \drawProportionalLine{BD,DC}^2$ (const.)
$\therefore \mbox{\inax[ax:II] } \squareCBFE = \drawFromCurrentPicture[bottom]{draw byNamedPolygon(DCLH,CLKA,HLEG) rotated 90;}$
$\therefore \drawProportionalLine{BD} \cdot \drawProportionalLine{DC,CA} + \drawProportionalLine{DC}^2 = \drawProportionalLine{CA}^2 = \drawProportionalLine{BD,DC}^2$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. VI. theor.},reference=prop:II.VI]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, K, L, M;
numeric h, s;
h := 5u;
s := 2/5h;
A := (0, h);
B := (0, 0);
C := A shifted (0, -s);
D := A shifted (0, -2s);
E := (-h+s, h-s);
F := (-h+s, 0);
G := (xpart(F), ypart(D));
H = whatever[D, G] = whatever[B, E];
K := (xpart(H), ypart(A));
L := (xpart(H), ypart(C));
M := (xpart(H), ypart(B));
draw byPolygon(B,D,H,M)(byblue);
draw byPolygon(D,C,L,H)(byyellow);
draw byPolygon(C,L,K,A)(black);
draw byPolygon(M,H,G,F)(byyellow);
draw byPolygon(H,L,E,G)(byred);
draw byLine(H, G, byblue, 1, 0);
draw byLine(D, H, byred, 0, 0);
byLineDefine(L, M, black, 1, 0);
byLineDefine(C, L, byred, 0, 0);
byLineDefine(B, D, byred, 0, 0);
byLineDefine(D, C, byblue, 0, 0);
byLineDefine(C, A, byyellow, 0, 0);
byLineDefine(E, B, black, 0, 0);
byLineDefine(A, K, byred, 1, 0);
byLineDefine(K, L, byyellow, 0, 0);
byLineDefine(L, E, black, 1, 0);
draw byNamedLineSeq(-1)(BD,DC,CA,AK,KL,LM,CL,LE,EB);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a straight line be bisected \drawProportionalLine{DC,CA} and produced to any point \drawProportionalLine{BD,DC,CA}, the rectangle contained by the whole line so increased and the part produced, together with the square of half the line, is equal to the square of the line made up of the half, and the produced part.\\
$\drawProportionalLine{BD,DC,CA} \cdot \drawProportionalLine{BD} + \drawProportionalLine{DC}^2 = \drawProportionalLine{BD,DC}^2$
}
\startCenterAlign
Describe \drawPolygon[bottom][squareCBFE]{BDHM,DCLH,MHGF,HLEG} \inprop[prop:I.XLVI], draw \drawProportionalLine{EB}\\
and $\left\{\eqalign{
\drawProportionalLine{HG,DH} & \parallel \drawProportionalLine{CL,LE} \cr
\drawProportionalLine{LM,KL} & \parallel \drawProportionalLine{BD,DC,CA} \cr
\drawProportionalLine{AK} & \parallel \drawProportionalLine{CL,LE}
}\right\}$ \inprop[prop:I.XXXI]
$\drawPolygon[bottom]{MHGF} = \drawPolygon[bottom]{DCLH} = \drawPolygon[bottom]{CLKA}$ (\inpropL[prop:I.XXXVI], \inpropL[prop:I.XLIII])
$\therefore \drawPolygon[bottom]{BDHM,DCLH,MHGF} = \drawPolygon[bottom]{BDHM,DCLH,CLKA} = \drawProportionalLine{BD} \cdot \drawProportionalLine{BD,DC,CA}$;\\
but $\drawPolygon[bottom]{HLEG} = \drawProportionalLine{DC}^2$ \inprop[prop:II.IV]
$\therefore \squareCBFE = \drawProportionalLine{HG,DH}^2 = \drawPolygon[bottom]{BDHM,DCLH,CLKA,HLEG}$ (const, \inaxL[ax:II])
$\therefore \drawProportionalLine{BD,DC,CA} \cdot \drawProportionalLine{BD} + \drawProportionalLine{DC}^2 = \drawProportionalLine{BD,DC}^2$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. VII. theor.},reference=prop:II.VII]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, N;
numeric w;
w := 7/2u;
A := (0, w);
B := (w, w);
C := 3/5[A, B];
D := (0, 0);
E := (w, 0);
N := 3/5[D, E];
F := 3/5[E, B];
G = whatever[D, B] = whatever[N, C];
H := whatever[A, D] = whatever[F, G];
draw byPolygon(D,N,G,H)(byred);
draw byPolygon(N,E,F,G)(black);
draw byPolygon(H,G,C,A)(byyellow);
draw byPolygon(G,F,B,C)(byblue);
draw byLine(G, N, byblue, 0, 0);
draw byLine(G, F, byred, 0, 0);
draw byLine(G, H, black, 1, 0);
draw byLine(G, C, black, 1, 0);
byLineDefine(B, D, black, 0, 0);
byLineDefine(D, N, byblue, 0, 0);
byLineDefine(N, E, byred, 0, 0);
byLineDefine(E, B, byyellow, 0, 0);
draw byNamedLineSeq(-1)(BD,DN,NE,EB);
}
\drawCurrentPictureInMargin
\problemNP{I}{f}{a straight line be divided into any two parts \drawProportionalLine{DN,NE}, the squares of the whole line and one of the parts are equal to twice the rectangle contained by the whole line and that part, together with the square of the other parts.\\
$\drawProportionalLine{DN,NE}^2 + \drawProportionalLine{NE}^2 = 2\drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE} + \drawProportionalLine{DN}^2$
}
\startCenterAlign
Describe \drawPolygon[bottom][squareABED]{DNGH,NEFG,HGCA,GFBC}. \inprop[prop:I.XLVI], draw \drawProportionalLine{BD} \inpost[post:I],\\
and $\left\{\eqalign{
\drawProportionalLine{GN,GC} &\parallel \drawProportionalLine{EB} \cr
\drawProportionalLine{GH,GF} &\parallel \drawProportionalLine{DN,NE}
}\right\}$\\
$\drawPolygon[bottom]{HGCA} = \drawPolygon[bottom]{NEFG}$ \inprop[prop:I.XLIII],\\
add $\drawPolygon[bottom]{GFBC} = \drawProportionalLine{NE}^2$ to both \inprop[prop:II.IV]
$\drawPolygon[bottom]{HGCA,GFBC} = \drawPolygon[bottom]{NEFG,GFBC} = \drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE}$\\
$\drawPolygon[bottom]{DNGH} = \drawProportionalLine{DN}^2$ \inprop[prop:II.IV]\\
$\drawPolygon[bottom]{HGCA,GFBC} + \drawPolygon[bottom]{NEFG,GFBC} + \drawPolygon[bottom]{DNGH} = 2\drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE} + \drawProportionalLine{DN}^2 = \squareABED + \drawPolygon[bottom]{GFBC}$;
$\drawProportionalLine{DN,NE}^2 + \drawProportionalLine{NE}^2 = 2\drawProportionalLine{DN,NE} \cdot \drawProportionalLine{NE} + \drawProportionalLine{DN}^2$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. VIII. theor.},reference=prop:II.VIII]
\defineNewPicture{
pair A, B, C, D, E, F, G, H, K, L, M, N, O, P, Q, R;
numeric w, d;
w := 7/2u;
d := u;
A := (0, w + d);
B := (w, w + d);
C := (w - d, w + d);
D := (w + d, w + d);
E := (0, 0);
F := (w + d, 0);
G := (w - d, w);
H := (w - d, 0);
K := (w, w);
L := (w, 0);
M := (0, w);
N := (w + d, w);
O := (0, w - d);
P := (w + d, w - d);
Q := (w - d, w - d);
R := (w, w - d);
draw byLine(C, Q, byblue, 1, 0);
draw byLine(B, R, black, 1, 0);
draw byLine(Q, H, byblue, 0, 0);
draw byLine(R, L, byblue, 0, 0);
draw byLine(M, G, black, 1, 0);
draw byLine(O, Q, byred, 1, 0);
draw byLine(G, N, byred, 0, 0);
draw byLine(Q, P, byred, 0, 0);
draw byLine(D, E, black, 0, 0);
byLineDefine(E, H, byblue, 0, 0);
byLineDefine(H, L, byred, 0, 0);
byLineDefine(L, F, byyellow, 0, 0);
byLineDefine(F, P, byblue, 0, 0);
byLineDefine(P, N, byyellow, 0, 0);
byLineDefine(N, D, byred, 0, 0);
byLineDefine(A, D, black, 0, 0);
byLineDefine(E, A, black, 0, 0);
draw byNamedLineSeq(0)(EH,HL,LF,FP,PN,ND,AD,EA);
byLineDefine(D, F, black, 0, 0);
byLineDefine(B, L, black, 0, 1);
byLineDefine(C, H, black, 0, 1);
byLineDefine(M, N, black, 0, 1);
byLineDefine(O, P, black, 0, 1);
}
\drawCurrentPicture
\problemNP{I}{f}{a straight line be divided into any two parts \drawProportionalLine{EH,HL}, the square of the sum of the whole line and any of its parts is equal to four times the rectangle contained by the whole line, and that part together with the square of the other part. \\
$\drawProportionalLine{EH,HL,LF}^2 = 4 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$
}
\startCenterAlign
Produce \drawProportionalLine{EH,HL} and make $\drawProportionalLine{LF} = \drawProportionalLine{HL}$
Construct \drawFromCurrentPicture[bottom]{
draw byNamedLine(BL,CH,MN,OP);
draw byNamedLineSeq(0)(LF,HL,EH,EA,AD,DF);
} \inprop[prop:I.XLVI];\\
draw \drawProportionalLine{DE},
$\left.\eqalign{
\left.\eqalign{
\vcenter{
\nointerlineskip\hbox{\drawProportionalLine{CQ,QH}}
\nointerlineskip\hbox{\drawProportionalLine{BR,RL}}
}
}\right\} & \parallel \drawProportionalLine{FP,PN,ND}\cr
\left.\eqalign{
\vcenter{
\nointerlineskip\hbox{\drawProportionalLine{OQ,QP}}
\nointerlineskip\hbox{\drawProportionalLine{MG,GN}}
}
}\right\} & \parallel \drawProportionalLine{EH,HL,LF}\cr
}\right\}$ \inprop[prop:I.XXXI]
$\drawProportionalLine{EH,HL,LF}^2 = \drawProportionalLine{LF}^2 + \drawProportionalLine{EH,HL}^2 + 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{LF}$ \inprop[prop:II.IV]\\
but $\drawProportionalLine{HL}^2 + \drawProportionalLine{EH,HL}^2 = 2 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$ \inprop[prop:II.VII]
$\therefore \drawProportionalLine{EH,HL,LF}^2 = 4 \cdot \drawProportionalLine{EH,HL} \cdot \drawProportionalLine{HL} + \drawProportionalLine{EH}^2$
\stopCenterAlign
\qed
\stopProposition
\startProposition[title={Prop. IX. theor.},reference=prop:II.IX]
\defineNewPicture{
pair A, B, C, D, E, F, G, H;
numeric w;
w := 5u;
A := (-1/2w, 0);
B := (1/2w, 0);
C := (0, 0);
D := (1/5w, 0);
E := (0, 1/2w);
F = whatever[E, B] = (xpart(D), whatever);
G = whatever[E, C] = (whatever, ypart(F));
H := 1/2[E, F];
draw byAngleWithName(E, A, B, byyellow, 0)(A);
draw byAngleWithName(A, B, E, byblue, 0)(B);
draw byAngle(A, E, C, byyellow, 0);
draw byAngle(C, E, B, byred, 0);
draw byAngle(E, F, G, byred, 0);
draw byAngle(D, F, B, black, 0);
draw byLine(A, F, black, 0, 0);