randomTriangleAngles.triangle() [[0, 1], []] [[1], [2]] function(){ var trA = new Triangle([5, -8], ANGLES, 14, {}); trA.boxOut([[[0, -10], [0, 10]]], [0.4, 0]); trA.boxOut([[[11, -10], [11, 10]]], [-0.4, 0]); trA.color = "blue"; return trA; }() function(){ var pointD = findIntersection(bisectAngle(TR_A.sides[0], reverseLine(TR_A.sides[2]), 1), TR_A.sides[1]); return pointD; }() function(){ var trB = new Triangle([0, 0], [], 3, {}, [TR_A.points[0], TR_A.points[1], POINT_D]); trB.labels = { "sides" : mergeArray( clearArray( trB.niceSideLengths, SIDES_B[ 0 ] ), clearArray( [ "?", "?", "?" ], SIDES_B[ 1 ] ) ), "points": [ "A", "B", "D" ] }; return trB; }() function(){ var trC = new Triangle( [0,0], [], 3, {}, [ TR_A.points[ 0 ], POINT_D, TR_A.points[ 2 ] ] ); trC.labels = { "angles" : clearArray( trC.niceAngles, [] ), "sides" : mergeArray( clearArray( trC.niceSideLengths, SIDES_C[ 0 ] ), clearArray( [ "?", "?", "?" ], SIDES_C[ 1 ] ) ), "points": [ "", "", "C" ] }; return trC; }() TR_B.niceSideLengths[ 1 ] TR_B.niceSideLengths[ 0 ] TR_C.niceSideLengths[ 1 ] TR_C.niceSideLengths[ 2 ] TR_B.niceSideLengths[ 1 ] TR_B.niceSideLengths[ 0 ] TR_C.niceSideLengths[ 1 ] TR_C.niceSideLengths[ 2 ]
var aBounding = TR_A.boundingRange(1); var bBounding = TR_D.boundingRange(1); var minX = Math.min(aBounding[0][0], bBounding[0][0]); var maxX = Math.max(aBounding[0][1], bBounding[0][1]); var minY = Math.min(aBounding[1][0], bBounding[1][0]); var maxY = Math.max(aBounding[1][1], bBounding[1][1]); init({ range: [[minX, maxX], [minY, maxY]] }) // draw an arc representing congruent angles BAD, CAD style({ stroke: GREEN }); var angle_mod = ((180 - TR_A.angles[0]) / 135) + 0.2; arc(TR_A.points[0], angle_mod, 0, TR_A.angles[0]); line([(angle_mod - 0.1) * cos(3 * TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][0], (angle_mod - 0.1) * sin(3 * TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][1]], [ (angle_mod + 0.1) * cos(3 * TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][0], (angle_mod + 0.1) * sin(3 * TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][1]]); line([(angle_mod - 0.1) * cos(TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][0], (angle_mod - 0.1) * sin(TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][1]], [ (angle_mod + 0.1) * cos(TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][0], (angle_mod + 0.1) * sin(TR_A.angles[0] / 4 * PI / 180) + TR_A.points[0][1]]); style({ stroke: "black" }); TR_B.draw(); TR_B.drawLabels(); TR_C.draw(); TR_C.drawLabels(); // draw a pink line over the line segment in question line(QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: PINK, strokeWidth: 3 });
roundTo(1, TEMP_AB * TEMP_CD / TEMP_BD) TR_A.points[0] TR_A.points[2] function(){ var pA = TR_A.points[0]; var pD = POINT_D; var pointE = [(pD[0] + (pD[0] - pA[0]) * (BD / CD)), (pD[1] + (pD[1] - pA[1]) * (BD / CD))]; return pointE; }() function(){ var trD = new Triangle([0, 0], [], 3, {}, [POINT_D, TR_A.points[1], POINT_E]); trD.labels = { "points": ["", "", "E"] }; return trD; }()

What is the length of side AC? (Round to 1 decimal place)

AC

There aren't any similar triangles in the problem figure, but we can make a new triangle with a few lines.
1. Draw a line parallel to the line \pink{AC}, through the point B.
2. Extend the line AD out to meet it at a new point E.

style({ stroke: "purple" }); TR_D.draw(); TR_D.drawLabels();

Now we have a very useful triangle \triangle BDE which is similar to \triangle ACD.

Since \pink{AC} and BE are parallel, we know that \angle CAE is congruent to \angle AEB.

style({ stroke: GREEN }); var angle_mod_d = ((180 - TR_D.angles[2]) / 180) + 0.2; arc(TR_D.points[2], angle_mod_d, 180 + TR_D.angles[2], 180 + TR_D.angles[2] * 2); line([(angle_mod_d - 0.1) * cos((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][0], (angle_mod_d - 0.1) * sin((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][1]], [(angle_mod_d + 0.1) * cos((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][0], (angle_mod_d + 0.1) * sin((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][1]]);

We can now tell that \text{m}\angle BAE is equal to \text{m}\angle AEB. Meaning \triangle ABE is isosceles, and we know the length of BE.

line(TR_A.points[0], TR_A.points[1], { stroke: BLUE, strokeWidth: 3 }); line(TR_A.points[1], POINT_E, { stroke: BLUE, strokeWidth: 3 }); TR_D.labels = { "sides": ["", AB, ""] }; TR_D.drawLabels();

\dfrac{AB}{BD} = \dfrac{\pink{AC}}{CD}

\pink{AC} = \dfrac{AB \cdot CD}{BD}

\pink{AC} = AC

roundTo(1, TEMP_AC * TEMP_BD / TEMP_CD) [[1], [0]] [[1, 2], []] TR_A.points[0] TR_A.points[1] function(){ var pA = TR_A.points[0]; var pD = POINT_D; var pointE = [(pD[0] + (pD[0] - pA[0]) * (CD / BD)), (pD[1] + (pD[1] - pA[1]) * (CD / BD))]; return pointE; }() function(){ var trD = new Triangle([0, 0], [], 3, {}, [POINT_D, TR_A.points[2], POINT_E]); trD.labels = { "points": ["", "", "E"] }; return trD; }()

What is the length of side AB? (Round to 1 decimal place)

AB

There aren't any similar triangles in the problem figure, but we can make a new triangle with a few lines.
1. Draw a line parallel to the line \pink{AB}, through the point C.
2. Extend the line AD out to meet it at a new point E.

style({ stroke: "purple" }); TR_D.draw(); TR_D.drawLabels();

Now we have a very useful triangle \triangle CDE which is similar to \triangle ABD.

Since \pink{AB} and CE are parallel, we know that \angle BAE is congruent to \angle AEB.

style({ stroke: GREEN }); var angle_mod_d = ((180 - TR_D.angles[2]) / 180) + 0.2; arc(TR_D.points[2], angle_mod_d, 180, 180 + TR_D.angles[2]); line([(angle_mod_d - 0.1) * cos((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][0], (angle_mod_d - 0.1) * sin((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][1]], [(angle_mod_d + 0.1) * cos((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][0], (angle_mod_d + 0.1) * sin((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][1]]);

We can now tell that \text{m}\angle CAE is equal to \text{m}\angle AEC. Meaning \triangle ACE is isosceles, and we know the length of \blue{CE}.

line(TR_A.points[0], TR_A.points[2], { stroke: BLUE, strokeWidth: 3 }); line(TR_A.points[2], POINT_E, { stroke: BLUE, strokeWidth: 3 }); TR_D.labels = { "sides": ["", "", AC] }; TR_D.drawLabels();

\dfrac{\pink{AB}}{BD} = \dfrac{AC}{CD}

\pink{AB} = \dfrac{ AC \cdot BD }{ CD }

\pink{AB} = AB

roundTo(1, TEMP_AC * TEMP_BD / TEMP_AB) [[1, 0], []] [[2], [1]] TR_A.points[2] POINT_D function(){ var pA = TR_A.points[0]; var pD = POINT_D; var pointE = [(pD[0] + (pD[0] - pA[0]) * (CD / BD)), (pD[1] + (pD[1] - pA[1]) * (CD / BD))]; return pointE; }() function(){ var trD = new Triangle([0,0], [], 3, {}, [POINT_D, TR_A.points[2], POINT_E]); trD.labels = { "points": ["", "", "E"] }; return trD; }()

What is the length of side CD? (Round to 1 decimal place)

CD

There aren't any similar triangles in the problem figure, but we can make a new triangle with a few lines.
1. Draw a line parallel to the line AB, through the point C.
2. Extend the line AD out to meet it at a new point E.

style({ stroke: "purple" }); TR_D.draw(); TR_D.drawLabels(); // redraw the pink line for the problem line(QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: PINK, strokeWidth: 3 });

Now we have a very useful triangle \triangle CDE which is similar to \triangle ABD.

Since AB and CE are parallel, we know that \angle BAE is congruent to \angle AEC.

style({ stroke: GREEN }); var angle_mod_d = ((180 - TR_D.angles[2]) / 180) + 0.2; arc( TR_D.points[2], angle_mod_d, 180, 180 + TR_D.angles[2] ); line([(angle_mod_d - 0.1) * cos((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][0], (angle_mod_d - 0.1) * sin((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][1]], [(angle_mod_d + 0.1) * cos((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][0], (angle_mod_d + 0.1) * sin((180 + TR_D.angles[2] / 2) * PI/180) + TR_D.points[2][1]]);

We can now see that \text{m}\angle CAE is equal to \text{m}\angle AEC. Meaning \triangle ACE is isosceles, and we know the length of \blue{CE}.

line(TR_A.points[0], TR_A.points[2], { stroke: BLUE, strokeWidth: 3 }); line(TR_A.points[2], POINT_E, { stroke: BLUE, strokeWidth: 3 }); TR_D.labels = { "sides": ["", AC, ""] }; TR_D.drawLabels();

\dfrac{ AB}{BD} = \dfrac{AC}{\pink{CD}}

\pink{CD} = \dfrac{AC \cdot BD}{AB}

\pink{CD} = CD

roundTo(1, TEMP_AB * TEMP_CD / TEMP_AC) [[0], [1]] [[1, 2], []] TR_A.points[1] POINT_D function(){ var pA = TR_A.points[0]; var pD = POINT_D; var pointE = [(pD[0] + (pD[0] - pA[0]) * (BD / CD)), (pD[1] + (pD[1] - pA[1]) * (BD / CD))]; return pointE; }() function(){ var trD = new Triangle([0,0], [], 3, {}, [POINT_D, TR_A.points[1], POINT_E]); trD.labels = { "points": ["", "", "E"] }; return trD; }()

What is the length of the side BD? (Round to 1 decimal place).

BD

There aren't any similar triangles in the problem figure, but we can make a new triangle with a few lines.
1. Draw a line parallel to the line AC, through the point B.
2. Extend the line AD out to meet it at a new point E.

style({ stroke: "purple" }); TR_D.draw(); TR_D.drawLabels(); line(QUESTION_POINT_Q, QUESTION_POINT_R, { stroke: PINK, strokeWidth: 3 });

Now we have a very useful triangle \triangle BDE which is similar to \triangle ACD.

Since AC and BE are parallel, we know that \angle CAE is congruent to \angle AEB.

style({ stroke: GREEN }); var angle_mod_d = ((180 - TR_D.angles[2]) / 180) + 0.2; arc(TR_D.points[2], angle_mod_d, 180 + TR_D.angles[2], 180 + TR_D.angles[2] * 2); line([(angle_mod_d - 0.1) * cos((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][0], (angle_mod_d - 0.1) * sin((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][1]], [(angle_mod_d + 0.1) * cos((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][0], (angle_mod_d + 0.1) * sin((180 + TR_D.angles[2] * 1.5) * PI/180) + TR_D.points[2][1]]);

We can now tell that \text{m}\angle BAE is equal to \text{m}\angle AEB. Meaning \triangle ABE is isosceles, and we know the length of \blue{BE}.

line(TR_A.points[0], TR_A.points[1], { stroke: BLUE, strokeWidth: 3 }); line(TR_A.points[1], POINT_E, { stroke: BLUE, strokeWidth: 3 }); TR_D.labels = { "sides": ["", AB, ""] }; TR_D.drawLabels();

\dfrac{AB}{\pink{BD}} = \dfrac{AC}{CD}

\pink{BD} = \dfrac{AB \cdot CD}{AC}

\pink{BD} = BD