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constructions_1.html
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constructions_1.html
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<!DOCTYPE html>
<html data-require="math graphie graphie-geometry interactive constructions matrix">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Constructions 1</title>
<script src="../khan-exercise.js"></script>
<script>
var applyRefFrame = function(coord, rot) {
var rotCoord = [];
rotCoord[0] = Math.cos(rot) * coord[0] - Math.sin(rot) * coord[1];
rotCoord[1] = Math.sin(rot) * coord[0] + Math.cos(rot) * coord[1];
return rotCoord;
};
</script>
</head>
<body>
<div class="exercise">
<div class="vars">
<var id="ROT">
random() < 0.5 ?
randRange(10, 80) * PI / 180 :
randRange(100, 170) * PI / 180
</var>
<var id="SLOPE">tan(ROT)</var>
<var id="HGRAPH">0</var>
</div>
<div class="problems">
<div id="perpendicular">
<div class="problem">
<form>
<input type="button" value="Add Compass"
onClick="javascript:
KhanUtil.construction.addCompass();
" />
<input type="button" value="Add Straightedge"
onClick="javascript:
KhanUtil.construction.addStraightedge();
" />
<input type="button" value="Clear"
onClick="javascript:
KhanUtil.construction.removeAllTools();
" />
</form>
<p class="question">
Construct a line perpendicular to the given line.
</p>
<div class="graphie" id="construction">
init({
range: [[-5, 5], [-5, 5]],
scale: 50
});
addMouseLayer();
addConstruction("construction");
var a = applyRefFrame([4, 0], ROT);
var b = applyRefFrame([-4, 0], ROT);
addDummyStraightedge(a, b);
</div>
</div>
<div class="solution" data-type="custom">
<div class="instruction">
Use the compass and straightedge tools to construct
a line perpendicular to the given line.
</div>
<div class="guess">
_.map(_.filter(construction.tools, function(tool) {
return tool.dummy !== true;
}), function(tool) {
if (tool.first != null) {
return {
first: {
coord: [
tool.first.coord[0],
tool.first.coord[1]
]
},
second: {
coord: [
tool.second.coord[0],
tool.second.coord[1]
]
}
};
} else if (tool.center != null) {
return {
center: {
coord: [
tool.center.coord[0],
tool.center.coord[1]
]
},
radius: tool.radius
};
}
})
</div>
<div class="validator-function">
// If there's only one element, it's the given line
if (guess.length === 0) {
return "";
}
// first determine if there is a perp. line
perp = null;
_.each(guess, function(tool) {
if (tool.first != null) {
ang = atan2(
tool.second.coord[1] - tool.first.coord[1],
tool.second.coord[0] - tool.first.coord[0]);
deg = ang * 180 / PI;
origDeg = ROT * 180 / PI;
if (abs(deg-origDeg+90) < 7 ||
abs(deg-origDeg-90) < 7) {
perp = tool;
}
}
});
if (perp == null) {
return false;
}
// next make sure there are two compasses,
// each of which are centered on the line
comps = _.filter(guess, function(tool) {
return tool.center != null;
});
if (comps.length < 2) {
return false;
}
onLine = [];
_.each(comps, function(comp1) {
_.each(comps, function(comp2) {
thisSlope = (comp1.center.coord[1] -
comp2.center.coord[1]) / (comp1.center.coord[0]
- comp2.center.coord[0]);
thisYInt = comp1.center.coord[1] -
thisSlope*comp1.center.coord[0];
if (abs(thisSlope - SLOPE) < 0.5 &&
abs(thisYInt) < 0.1) {
onLine = [comp1,comp2];
}
});
});
// Really, Javascript? [] !== []? Fine.
if (onLine.length < 2) {
return false;
}
// now we know that the slope of the straightedge
// is good, and the two compasses are an on the line,
// so if the straightedge has the same y-intercept
// as the y-intercept of the line going between
// the two points of intersection of the two compasses
var a = onLine[0].center.coord[0];
var b = onLine[0].center.coord[1];
var c = onLine[1].center.coord[0];
var d = onLine[1].center.coord[1];
var r = onLine[0].radius;
var s = onLine[1].radius;
var e = c - a;
var f = d - b;
var p = sqrt(pow(e,2) + pow(f,2));
var k = (pow(p,2) + pow(r,2) -
pow(s,2))/(2*p);
var x1 = a + e * k / p + (f / p)
* sqrt(pow(r, 2) - pow(k, 2));
var y1 = b + f * k / p - (e / p)
* sqrt(pow(r, 2) - pow(k, 2));
interYInt = y1 + x1 * (1 / SLOPE);
perpYInt = perp.first.coord[1]
+ perp.first.coord[0] * (1 / SLOPE);
// give some leeway for the y-int when the slope
// is high
return abs(interYInt - perpYInt) < 1;
</div>
<div class="show-guess">
showConstructionGuess(guess);
</div>
</div>
<div class="hints">
<div>
<div class="graphie" data-update="construction">
graph.perp = raphael.set();
graph.perp.push(line(
applyRefFrame([0, 10], ROT),
applyRefFrame([0, -10], ROT), {
strokeWidth: 1,
stroke: BLUE
})).toBack();
</div>
<p>
We could just draw a line and <em>try</em> to make it
perpendicular, but then <strong>we have no guarantee
</strong> that it's actually perfectly perpendicular.
</p>
<p>
How can you guarantee that a line is perpendicular?
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.perpPoints = raphael.set();
graph.hintLines = raphael.set();
style({
fill: BLUE,
stroke: null,
}, function() {
graph.perpPoints.push(circle(
applyRefFrame([0, -1], ROT), 0.08));
graph.perpPoints.push(circle(
applyRefFrame([0, 1], ROT), 0.08));
});
graph.perp.push(drawHintLine(
applyRefFrame([0, 0], ROT),
applyRefFrame([0, 1], ROT), 1));
graph.perp.push(drawHintLine(
applyRefFrame([0, 0], ROT),
applyRefFrame([0, -1], ROT), 1));
graph.hintLines.push(drawHintLine(
applyRefFrame([0, -1], ROT),
applyRefFrame([1, 0], ROT), 2));
graph.hintLines.push(drawHintLine(
applyRefFrame([0, 1], ROT),
applyRefFrame([1, 0], ROT), 2));
graph.hintLines.push(drawHintLine(
applyRefFrame([0, -1],ROT),
applyRefFrame([-2, 0],ROT), 3));
graph.hintLines.push(drawHintLine(
applyRefFrame([0, 1],ROT),
applyRefFrame([-2, 0],ROT), 3));
graph.perp.toBack();
graph.hintLines.toBack();
graph.perpPoints.toBack();
</div>
<p>
If we pick <span class="hint_blue">
two points</span> on the perpendicular line
which are an equal distance from the intersection,
they will also be the same distance from every other
point on the line we started with.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.perp.remove();
</div>
<p>
If we don't already have the perpendicular line, is there
another way to find the blue points?
</p>
</div>
<div>
<div class="graphie" data-update="construction">
circle(applyRefFrame([1, 0], ROT), 0.08, {
fill: GRAY,
stroke: null
});
circle(applyRefFrame([1, 0], ROT),
eDist(applyRefFrame([0, -1], ROT),
applyRefFrame([1, 0], ROT)), {
stroke: GRAY,
strokeWidth: 1,
fill: "none",
strokeDasharray: "- "
});
</div>
<p>
If we use the compass to put a circle somewhere on the
line, the circle will include all points that are the
same distance from that point, including the two blue
points.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
circle(applyRefFrame([-2, 0], ROT), 0.08, {
fill: GRAY,
stroke: null
});
circle(applyRefFrame([-2, 0], ROT),
eDist(applyRefFrame([0, -1], ROT),
applyRefFrame([-2, 0], ROT)), {
stroke: GRAY,
strokeWidth: 1,
fill: "none",
strokeDasharray: "- "
});
</div>
<p>
We can add a second circle somewhere else on the line
that intersects with the first circle.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.hintLines.remove();
</div>
<p>
The points where the two circles intersect can be used
to draw a perpendicular line.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.perpPoints.attr({fill: GRAY});
line(applyRefFrame([0, 10], ROT),
applyRefFrame([0, -10], ROT), {
strokeWidth: 1,
stroke: GRAY,
strokeDasharray: "- "
}).toBack();
</div>
<p>
Use a straightedge to connect the two points where the
circles intersect. This line is perpendicular to the
given line.
</p>
</div>
</div>
</div>
<div id="bisector">
<div class="problem">
<form>
<input type="button" value="Add Compass"
onClick="javascript:
KhanUtil.construction.addCompass();
" />
<input type="button" value="Add Straightedge"
onClick="javascript:
KhanUtil.construction.addStraightedge();
" />
<input type="button" value="Clear"
onClick="javascript:
KhanUtil.construction.removeAllTools();
" \>
</form>
<p class="question">
Construct a perpendicular bisector of the line segment
<code>\overline{AB}</code>.
</p>
<div class="graphie" id="construction">
init({
range: [[-5,5],[-5,5]],
scale: 50
});
addMouseLayer();
addConstruction("construction");
var a = applyRefFrame([1.5, 0], ROT);
var b = applyRefFrame([-1.5, 0], ROT);
a = [roundToNearest(0.01, a[0]),
roundToNearest(0.01, a[1])];
b = [roundToNearest(0.01, b[0]),
roundToNearest(0.01, b[1])];
addDummyStraightedge(a, b, false);
addDummyPoint(a);
addDummyPoint(b);
var offset = "above";
if ((ROT * 180 / PI) > 50 &&
(ROT * 180 / PI) < 90) {
offset = "left";
}
if ((ROT * 180 / PI) < 130 &&
(ROT * 180 / PI) > 90) {
offset = "right";
}
label(a, "A", offset);
label(b, "B", offset);
</div>
</div>
<div class="solution" data-type="custom">
<div class="instruction">
Use the compass and straightedge tools to construct
a perpendicular bisector of <code>\overline{AB}</code>.
</div>
<div class="guess">
_.map(_.filter(construction.tools, function(tool) {
return tool.dummy !== true;
}), function(tool) {
if (tool.first != null) {
return {
first: {
coord: [
tool.first.coord[0],
tool.first.coord[1]
]
},
second: {
coord: [
tool.second.coord[0],
tool.second.coord[1]
]
}
};
} else if (tool.center != null) {
return {
center: {
coord: [
tool.center.coord[0],
tool.center.coord[1]
]
},
radius: tool.radius
};
}
})
</div>
<div class="validator-function">
// If there's only one element, it's the given line
if (guess.length === 0) {
return "";
}
// first determine if there is a perp. line
perp = null;
_.each(guess, function(tool) {
if (tool.first != null) {
ang = atan2(
tool.second.coord[1] - tool.first.coord[1],
tool.second.coord[0] - tool.first.coord[0]);
deg = ang * 180 / PI;
origDeg = ROT * 180 / PI;
if (abs(deg - origDeg + 90) < 7 ||
abs(deg - origDeg - 90) < 7) {
perp = tool;
}
}
});
if (perp == null) {
return false;
}
// next make sure there are two compasses,
// each of which are centered on the line,
// with equal radius
comps = _.filter(guess, function(tool) {
return tool.center != null;
});
if (comps.length < 2) {
return false;
}
onLine = [];
_.each(comps, function(comp1) {
_.each(comps, function(comp2) {
thisSlope = (comp1.center.coord[1]
- comp2.center.coord[1])
/ (comp1.center.coord[0]
- comp2.center.coord[0]);
thisYInt = comp1.center.coord[1]
- thisSlope*comp1.center.coord[0];
if (abs(thisSlope - SLOPE) < 0.5 &&
abs(thisYInt) < 0.1 &&
abs(comp1.radius - comp2.radius) < 0.1) {
onLine = [comp1,comp2];
}
});
});
// Really, Javascript? [] !== []? Fine.
if (onLine.length < 2) {
return false;
}
// now we know that the slope of the straightedge
// is good, and the two compasses are an on the line,
// so if the straightedge has the same y-intercept
// as the y-intercept of the line going between
// the two points of intersection of the two compasses
interYInt = 0;
perpYInt = perp.first.coord[1]
+ perp.first.coord[0]*(1/SLOPE);
return abs(interYInt - perpYInt) < 0.5;
</div>
<div class="show-guess">
showConstructionGuess(guess);
</div>
</div>
<div class="hints">
<div>
<div class="graphie" data-update="construction">
graph.perp = raphael.set();
graph.perp.push(line(
applyRefFrame([0, 10], ROT),
applyRefFrame([0, -10], ROT), {
strokeWidth: 1,
stroke: BLUE
})).toBack();
</div>
<p>
We could just draw a line and <em>try</em> to get it
right, but then <strong>we have no guarantee</strong>
that it's actually perfectly perpendicular or that it
bisects the segment at exactly the midpoint.
</p>
<p>
How can you guarantee that a line is really a
perpendicular bisector?
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.hintLines = raphael.set();
style({
fill: BLUE,
stroke: null,
}, function() {
graph.perpPoint1 = circle(
applyRefFrame([0, -1], ROT), 0.08);
graph.perpPoint2 = circle(
applyRefFrame([0, 1], ROT), 0.08);
});
graph.perp.push(drawHintLine(
applyRefFrame([0, 0], ROT),
applyRefFrame([0, 1], ROT), 1));
graph.perp.push(drawHintLine(
applyRefFrame([0, 0], ROT),
applyRefFrame([0, -1], ROT), 1));
graph.hintLines.push(drawHintLine(
applyRefFrame([0, -1], ROT),
applyRefFrame([1.5, 0], ROT), 2));
graph.hintLines.push(drawHintLine(
applyRefFrame([0, 1], ROT),
applyRefFrame([1.5, 0], ROT), 2));
graph.hintLines.push(drawHintLine(
applyRefFrame([0, -1],ROT),
applyRefFrame([-1.5, 0],ROT), 2));
graph.hintLines.push(drawHintLine(
applyRefFrame([0, 1],ROT),
applyRefFrame([-1.5, 0],ROT), 2));
graph.perp.toBack();
graph.hintLines.toBack();
graph.perpPoint1.toBack();
graph.perpPoint2.toBack();
</div>
<p>
If we pick <span class="hint_blue">
two points</span> on the perpendicular bisector
which are an equal distance from the intersection,
they will also be the same distance from both
endpoints of the segment we started with.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.perp.remove();
</div>
<p>
If we don't already have the perpendicular bisector,
is there another way to find the blue points?
</p>
</div>
<div>
<div class="graphie" data-update="construction">
circle(applyRefFrame([1.5, 0], ROT), 0.08, {
fill: GRAY,
stroke: null
});
graph.compass1 = circle(applyRefFrame([1.5, 0], ROT),
eDist(applyRefFrame([0, -1], ROT),
applyRefFrame([1.5, 0], ROT)), {
stroke: GRAY,
strokeWidth: 1,
fill: "none",
strokeDasharray: "- "
}).toBack();
</div>
<p>
If we use the compass to put a circle centered at point
<code>A</code>, the circle will include all points that
are the same distance from point <code>A</code>,
including the two blue points.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
circle(applyRefFrame([-1.5, 0], ROT), 0.08, {
fill: GRAY,
stroke: null
});
graph.compass2 = circle(applyRefFrame([-1.5, 0], ROT),
eDist(applyRefFrame([0, -1], ROT),
applyRefFrame([-2, 0], ROT)), {
stroke: GRAY,
strokeWidth: 1,
fill: "none",
strokeDasharray: "- "
}).toBack();
</div>
<p>
We can add a second circle at point <code>B</code>
that intersects with the first circle.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.hintLines.remove();
</div>
<p>
But wait! We can use these circles to draw a
perpendicular line, <strong>but not a bisector!
</strong> That's because the two circles are different
sizes.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.compass1.animate({
rx: scaleVector(3)[0],
ry: scaleVector(3)[1]
}, 250);
graph.compass2.animate({
rx: scaleVector(3)[0],
ry: scaleVector(3)[1]
}, 250);
graph.perpPoint1.animate({
cx: scalePoint(applyRefFrame(
[0, -3/2 * sqrt(3)], ROT))[0],
cy: scalePoint(applyRefFrame(
[0, -3/2 * sqrt(3)], ROT))[1]
}, 250);
graph.perpPoint2.animate({
cx: scalePoint(applyRefFrame(
[0, 3/2 * sqrt(3)], ROT))[0],
cy: scalePoint(applyRefFrame(
[0, 3/2 * sqrt(3)], ROT))[1]
}, 250);
</div>
<p>
One nice way to make the circles the same size is to
set the radii equal to the distance between
<code>A</code> and <code>B</code>. You can do this by
setting the center at one point and the edge of the
circle at the other.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.perpPoint1.attr({fill: GRAY});
graph.perpPoint2.attr({fill: GRAY});
line(applyRefFrame([0, 10], ROT),
applyRefFrame([0, -10], ROT), {
strokeWidth: 1,
stroke: GRAY,
strokeDasharray: "- "
}).toBack();
</div>
<p>
Use a straightedge to connect the two points where the
circles intersect. This line is the perpendicular
bisector of <code>\overline{AB}</code>.
</p>
</div>
</div>
</div>
<div id="angle-bisect">
<div class="vars" data-apply="replace">
<var id="ROT">randRange(30, 80) * PI / 180</var>
<var id="CENTER">[randRange(-3, 0), 0]</var>
<var id="HGRAPH">1</var>
<var id="A">CENTER</var>
<var id="B">[CENTER[0] + 4, CENTER[1]]</var>
<var id="C">
(function() {
var c = applyRefFrame([4, 0], ROT);
c = [c[0] + CENTER[0], c[1] + CENTER[1]];
c = [roundToNearest(0.01, c[0]),
roundToNearest(0.01, c[1])];
return c;
})()
</var>
</div>
<div class="problem">
<form>
<input type="button" value="Add Compass"
onClick="javascript:
KhanUtil.construction.addCompass();
" />
<input type="button" value="Add Straightedge"
onClick="javascript:
KhanUtil.construction.addStraightedge();
" />
<input type="button" value="Clear"
onClick="javascript:
KhanUtil.construction.removeAllTools();
" \>
</form>
<p class="question">
Construct an angle bisector for the given angle.
</p>
<div class="graphie" id="construction">
init({
range: [[-5, 5], [-2, 5]],
scale: 50
});
addMouseLayer();
addConstruction("construction");
addDummyRay(A, B);
addDummyRay(A, C);
</div>
</div>
<div class="solution" data-type="custom">
<div class="instruction">
Use the compass and straightedge tools to construct
an angle bisector.
</div>
<div class="guess">
_.map(_.filter(construction.tools, function(tool) {
return tool.dummy !== true;
}), function(tool) {
if (tool.first != null) {
return {
first: {
coord: [
tool.first.coord[0],
tool.first.coord[1]
]
},
second: {
coord: [
tool.second.coord[0],
tool.second.coord[1]
]
}
};
} else if (tool.center != null) {
return {
center: {
coord: [
tool.center.coord[0],
tool.center.coord[1]
]
},
radius: tool.radius
};
}
})
</div>
<div class="validator-function">
if (guess.length === 0) {
return "";
}
// first determine if there is a line bisecting
var bisect = null;
// var guess = construction.tools;
_.each(guess, function(tool) {
if (tool.first != null) {
ang = atan2(
(tool.second.coord[1] - tool.first.coord[1]),
(tool.second.coord[0] - tool.first.coord[0]));
ang = ang < 0 ? ang + PI : ang;
if (abs((ROT / 2) - ang) < 3 * PI / 180) {
bisect = tool;
}
}
});
if (bisect == null) {
return false;
}
// next make sure there is a compass with
// center on the center of the angle such that
// its intersections with the angle are the
// centers of two further compasses (yup)
var middle = null;
var bottom = null;
var top = null;
var comps = _.filter(guess, function(tool) {
return tool.center != null;
});
if (comps.length < 3) {
return false;
}
_.each(comps, function(comp) {
if(eDist(comp.center.coord, CENTER) < 0.5) {
middle = comp;
}
});
if (middle == null) {
return false;
}
var botInter = [];
_.each(comps, function(comp) {
botInter = [CENTER[0] + comp.radius, CENTER[1]];
if (eDist(comp.center.coord, botInter) < 0.5) {
bottom = comp;
}
});
var topInter = [];
_.each(comps, function(comp) {
topInter = [CENTER[0] +
applyRefFrame([comp.radius, 0], ROT)[0], CENTER[1]
+ applyRefFrame([comp.radius, 0], ROT)[1]];
if( eDist(comp.center.coord, topInter) < 0.5) {
top = comp;
}
});
if (bottom == null || top == null) {
return false;
}
return true;
</div>
<div class="show-guess">
showConstructionGuess(guess);
</div>
</div>
<div class="hints">
<div>
<div class="graphie" data-update="construction">
var farPoint = [
applyRefFrame([10, 0], ROT / 2)[0] + CENTER[0],
applyRefFrame([10 , 0], ROT / 2)[1] + CENTER[1]
];
graph.bisect = line(CENTER, farPoint, {
stroke: BLUE,
strokeWidth: 1
});
graph.bisect.toBack();
</div>
<p>
We could just draw a line and <em>try</em> to make it
bisect the angle, but that's difficult to do
and <strong>there is no guarantee</strong> it's a
perfect bisector.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.hintLines = raphael.set();
graph.intersect1 = [CENTER[0] + 2, CENTER[1]];
graph.intersect2 = [
applyRefFrame([2, 0], ROT)[0] + CENTER[0],
applyRefFrame([2, 0], ROT)[1] + CENTER[1]];
circle(graph.intersect1, 0.08, { stroke: null, fill: BLUE });
circle(graph.intersect2, 0.08, { stroke: null, fill: BLUE });
graph.hintLines.push(drawHintLine(CENTER, graph.intersect1, 1));
graph.hintLines.push(drawHintLine(CENTER, graph.intersect2, 1));
graph.hintLines.push(drawHintLine(graph.intersect1, [
applyRefFrame([2.5, 0], ROT / 2)[0] + CENTER[0],
applyRefFrame([2.5, 0], ROT / 2)[1] + CENTER[1]], 2));
graph.hintLines.push(drawHintLine(graph.intersect2, [
applyRefFrame([2.5, 0], ROT / 2)[0] + CENTER[0],
applyRefFrame([2.5, 0], ROT / 2)[1] + CENTER[1]], 2));
graph.hintLines.push(drawHintLine(graph.intersect1, [
applyRefFrame([4, 0], ROT / 2)[0] + CENTER[0],
applyRefFrame([4, 0], ROT / 2)[1] + CENTER[1]], 3));
graph.hintLines.push(drawHintLine(graph.intersect2, [
applyRefFrame([4, 0], ROT / 2)[0] + CENTER[0],
applyRefFrame([4, 0], ROT / 2)[1] + CENTER[1]], 3));
graph.hintLines.toBack();
</div>
<p>
If we pick <span class="hint_blue">any two points on
the given lines</span> that are the same distance from
the vertex of the angle, every point on the bisector
line will be equidistant from those points.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
circle(CENTER, 0.08, {
stroke: null,
fill: GRAY
});
circle(CENTER, 2, {
fill: null,
stroke: GRAY,
strokeWidth: 1,
strokeDasharray: "- "
});
</div>
<p>
We can use a compass centered at the vertex to find
<span class="hint_blue">two points equidistant from
the vertex</span>.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.bisect.remove();
circle(graph.intersect1, 0.08, {
stroke: null,
fill: GRAY
});
graph.compass1 = circle(graph.intersect1, 1.5, {
fill: null,
stroke: GRAY,
strokeWidth: 1,
strokeDasharray: "- "
});
circle(graph.intersect2, 0.08, {
stroke: null,
fill: GRAY
});
graph.compass2 = circle(graph.intersect2, 1.8, {
fill: null,
stroke: GRAY,
strokeWidth: 1,
strokeDasharray: "- "
});
graph.hintLines.remove();
</div>
<p>
If we use two more compasses centered at each of the
<span class="hint_blue">two points</span>, we can see
that they intersect, <strong>but not on the angle
bisector!</strong> That's because the two circles are
not the same size.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
graph.compass1.animate({
rx: scaleVector(2)[0],
ry: scaleVector(2)[1]
}, 250);
graph.compass2.animate({
rx: scaleVector(2)[0],
ry: scaleVector(2)[1]
}, 250);
</div>
<p>
A nice way to make the circles the same size is to
set the edges of both circles so they pass through the
vertex.
</p>
</div>
<div>
<div class="graphie" data-update="construction">
circle([CENTER[0] + 2 + 2 * cos(ROT), CENTER[1] + 2 * sin(ROT)], 0.08, {
stroke: null,
fill: GRAY
});
line([
applyRefFrame([-10, 0], ROT / 2)[0] + CENTER[0],
applyRefFrame([-10 , 0], ROT / 2)[1] + CENTER[1]
], [
applyRefFrame([10, 0], ROT / 2)[0] + CENTER[0],
applyRefFrame([10 , 0], ROT / 2)[1] + CENTER[1]
], {
stroke: GRAY,
strokeWidth: 1,
strokeDasharray: "- "
});
</div>
<p>
Use a straightedge to connect the vertex to the point
where the last two circles intersect. This line is the
angle bisector.
</p>