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Geometry · Chapter 32

Similar Triangles

Same shape, different size: spot two equal angles and every length scales by k, every area by k². Plus the right-triangle altitude facts and the Angle Bisector Theorem.

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32.00%

Similar Triangles

In this section you will be able to spot similar triangles, scale lengths by k and areas by k², and use the altitude of a right triangle.

Same shape, different size: every length scales by kk and every area by k2k^2. Spot two equal angles and you get both for free.

Two triangles are similar when one is a scaled copy of the other. You rarely measure all three angles. Look for parallel lines (they make equal angles) and shared or vertical angles: two equal angles are enough.

Once you have them, write the ratios in matching order: the side opposite the same angle goes on top each time.

Right triangles hide three similar copies at once. Drop the altitude to the hypotenuse, and every product in Theorem 32.0.3 falls out of one ratio.

Exploreplay first, formulas later

Scale factor 2: every side is multiplied by 2, the area goes from 6 to 24, which is 4 times as much

scale factor k2
5341068smallbig (×2)
sides
× 2
perimeter
12 → 24
area
6 → 24 (× 4)
big sidesmall side=k=2,[big][small]=k2=4\frac{\text{big side}}{\text{small side}} = k = 2, \qquad \frac{[\text{big}]}{[\text{small}]} = k^2 = 4

k = 2: sides 5, 3, 4 become 10, 6, 8; area 6 becomes 24, ratio 4

Name it

Theorem 32.0.1 · Similar Triangles

For similar triangles:

  1. All the angles of the triangles are the same.
  2. All corresponding sides have the same ratio.
  3. The ratio of the areas is the square of the side length ratio.

DE parallel to BC with AD/AB = 1/2: DE = 4, area ratio 1/4

AD : AB1/2
4836ABCED
every side ratio
AD : AB = DE : BC = 1/2
area ratio
5.08 : 20.33 = 1/4

DE is parallel to BC. AD/AB = 1/2, DE = 4, area ratio 1/4

Concept 32.0.2 · Similarity Test

Two triangles are similar if the three angles in the triangles are the same. In other words, the triangles are the same shape multiplied by a scale factor. In general, triangles are similar if:

  • AA similarity: two angles of the triangles are the same (so the third angle is equal too).
  • SAS similarity (Side Angle Side): two sides are proportional and the angle between those sides is equal.
  • SSS similarity (Side Side Side): all three sides are proportional.
  • HL similarity (Hypotenuse Leg): in a right triangle, the hypotenuse and a leg are proportional.
  • LL similarity (Leg Leg): in a right triangle, the two legs are proportional.

Warning: SSA does not mean the triangles are similar.

An easy way to spot similar triangles: the bases of the triangles are parallel and their sides lie on the same lines (an hourglass, or one triangle sitting inside another).

AA: every triangle DEF with these data is similar to ABC

∠D = ∠A = 50°∠E = ∠B = 60°
ABCDEF
Same shape ✓ similarity test (scale factor 3/2)

AA: every triangle DEF with these data is similar to ABC

Theorem 32.0.3 · Special Properties of Right Triangles

In a right triangle ABCABC where BB is the right angle and BDBD is the altitude to the hypotenuse, the following triangles are similar:

△ABC∼△ADB∼△BDC.\triangle ABC \sim \triangle ADB \sim \triangle BDC.

The length of the perpendicular to the hypotenuse is

BD=AB⋅BCAC.BD = \frac{AB \cdot BC}{AC}.

Also:

AD⋅CD=BD2,AD⋅AC=AB2,CD⋅CA=CB2.AD \cdot CD = BD^2, \qquad AD \cdot AC = AB^2, \qquad CD \cdot CA = CB^2.

Right angle at B, BC = 3, AB = 4, AC = 5: AD = 16/5, CD = 9/5, BD = 12/5

leg a3
leg b4
ABCD3412/5
AC = 5, BD = 4·3 ÷ 5 = 12/5
  • AD·CD = 16/5 · 9/5 = 144/25 = BD²
  • AD·AC = 16/5 · 5 = 16 = AB²
  • CD·CA = 9/5 · 5 = 9 = CB²

Right angle at B, BC = 3, AB = 4, AC = 5: AD = 16/5, CD = 9/5, BD = 12/5

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Example 32.1

AIME

A point PP is chosen in the interior of △ABC\triangle ABC such that when lines are drawn through PP parallel to the sides of △ABC\triangle ABC, the resulting smaller triangles t1t_1, t2t_2, and t3t_3 in the figure have areas 44, 99, and 4949, respectively. Find the area of △ABC\triangle ABC.

Parallel slice (its total is revealed at the end of the example)

t1t2t3ABCP
corner triangles
t1t_1 = 4 · t2t_2 = 9 · t3t_3 = 49
square roots add up
[ABC] = ? (hidden until the end of the example)

Three lines through P parallel to the sides. Corner triangles t_1 = 4, t_2 = 9, t_3 = 49; whole triangle ?

  1. 1

    Each small triangle has its three sides parallel to the sides of △ABC\triangle ABC. How is it related to △ABC\triangle ABC?

    Predict first:
    Click a choice or type its letter (A–E).
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Example 32.2

AMC 10

In square ABCDABCD, points PP and QQ lie on AD‾\overline{AD} and AB‾\overline{AB}, respectively. Segments BP‾\overline{BP} and CQ‾\overline{CQ} intersect at right angles at RR, with BR=6BR = 6 and PR=7PR = 7. What is the area of the square?

  1. 1

    What is BPBP?

    Predict first:
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Example 32.3

AMC 10/12

Quadrilateral ABCDABCD satisfies ∠ABC=∠ACD=90∘\angle ABC = \angle ACD = 90^\circ, AC=20AC = 20, and CD=30CD = 30. Diagonals AC‾\overline{AC} and BD‾\overline{BD} intersect at point EE, and AE=5AE = 5. What is the area of quadrilateral ABCDABCD?

  1. 1

    Start with the easy half. What is [ACD][ACD]?

    Predict first:
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Try it

Problem 32.2.1

AMC 10

A paper triangle with sides of lengths 33, 44, and 55 inches, as shown, is folded so that point AA falls on point BB. What is the length in inches of the crease?

Fold A onto B: the crease DE is the perpendicular bisector of AB, length ?; DA = DB = ?; EA = EB = ?

DEACB
Crease DE = ? (hidden until the end of the example)DA = DB = ? · EA = EB = ?

Fold A onto B: the crease DE is the perpendicular bisector of AB, length ?; DA = DB = ?; EA = EB = ?

Problem 32.2.27(additional)

AMC 10

Rectangle ABCDABCD has AB=8AB = 8 and BC=6BC = 6. Point MM is the midpoint of diagonal AC‾\overline{AC}, and EE is on AB‾\overline{AB} with ME‾⊥AC‾\overline{ME} \perp \overline{AC}. What is the area of △AME\triangle AME?

Problem 32.2.29(additional)

AMC 10

In rectangle ABCDABCD, AB=5AB = 5 and BC=3BC = 3. Points FF and GG are on CD‾\overline{CD} so that DF=1DF = 1 and GC=2GC = 2. Lines AFAF and BGBG intersect at EE. Find the area of △AEB\triangle AEB.

Contest cornerreal contest problems

Problem 32.2.2

AMC 10

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of △ABC\triangle ABC?

Problem 32.2.4

AMC 10

Right triangle ABCABC has leg lengths AB=20AB = 20 and BC=21BC = 21. Including AB‾\overline{AB} and BC‾\overline{BC}, how many line segments with integer length can be drawn from vertex BB to a point on hypotenuse AC‾\overline{AC}?

Problem 32.2.6

AMC 10

In △ABC\triangle ABC with a right angle at CC, point DD lies in the interior of AB‾\overline{AB} and point EE lies in the interior of BC‾\overline{BC} so that AC=CDAC = CD, DE=EBDE = EB, and the ratio AC:DE=4:3AC : DE = 4 : 3. What is the ratio AD:DBAD : DB?

Problem 32.2.8

AMC 10

In rectangle ABCDABCD, AB=6AB = 6 and BC=3BC = 3. Point EE between BB and CC, and point FF between EE and CC are such that BE=EF=FCBE = EF = FC. Segments AE‾\overline{AE} and AF‾\overline{AF} intersect BD‾\overline{BD} at PP and QQ, respectively. The ratio BP:PQ:QDBP : PQ : QD can be written as r:s:tr : s : t where the greatest common factor of rr, ss, and tt is 11. What is r+s+tr + s + t?

Problem 32.2.12

AMC 10/12

The diagram below shows a rectangle with side lengths 44 and 88 and a square with side length 55. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

Problem 32.2.13

AMC 10B

Let ABCDABCD be a rhombus with ∠ADC=46∘\angle ADC = 46^\circ. Let EE be the midpoint of CD‾\overline{CD}, and let FF be the point on BE‾\overline{BE} such that AF‾\overline{AF} is perpendicular to BE‾\overline{BE}. What is the degree measure of ∠BFC\angle BFC?

Problem 32.2.24(additional)

AMC 10

Rectangle ABCDABCD has AB=3AB = 3 and BC=4BC = 4. Point EE is the foot of the perpendicular from BB to diagonal AC‾\overline{AC}. What is the area of △AED\triangle AED?

Checkpoint for section 32.0

Get 2 of 3 right to master this section. One try each, no hints.
1

Rectangle ABCDABCD has AB=6AB = 6 and BC=3BC = 3. Point MM is chosen on side AB‾\overline{AB} so that ∠AMD=∠CMD\angle AMD = \angle CMD. What is the degree measure of ∠AMD\angle AMD?

2

Two similar triangles have corresponding sides in the ratio 2:52 : 5. The smaller one has area 1212. What is the area of the larger one?

3

Right triangle ABCABC has its right angle at BB. The altitude BDBD meets the hypotenuse with AD=4AD = 4 and DC=9DC = 9. What is BDBD?

32.10%

Angle Bisector Theorem

In this section you will be able to use the Angle Bisector Theorem to split a side in the ratio of the two sides next to it.

A bisector splits the opposite side in the ratio of the two sides beside it. The longer side gets the longer piece.

The proof is about areas. Triangles ABDABD and ACDACD have the same height from AA, and DD is equally far from ABAB and ACAC. Compare the two areas both ways and the ratio drops out.

Two bisectors meet at the incenter. Use the theorem twice, and the incenter splits each bisector in the ratio (sum of the two sides at that vertex) : (opposite side).

Exploreplay first, formulas later

AB = 7.62, BD = 4.59, AC = 10.63, CD = 6.41: AB/BD = 1.66 and AC/CD = 1.66

D7.6210.634.596.41ABC

Drag A, B or C. The two halves of angle A stay equal: 36° and 36°.

7.62 ÷ 4.59 = 1.66 and 10.63 ÷ 6.41 = 1.66

AB = 7.62, BD = 4.59, AC = 10.63, CD = 6.41: AB/BD = 1.66 and AC/CD = 1.66

Name it

Theorem 32.1.1 · Angle Bisector Theorem

If the line ADAD bisects angle AA of triangle ABCABC (with DD on BCBC), then

ABBD=ACCD.\frac{AB}{BD} = \frac{AC}{CD}.

Equivalently, BD:DC=AB:ACBD : DC = AB : AC: the longer side gets the longer piece.

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Example 32.4

AMC 12

In △ABC\triangle ABC, AB=6AB = 6, BC=7BC = 7, and CA=8CA = 8. Point DD lies on BC‾\overline{BC}, and AD‾\overline{AD} bisects ∠BAC\angle BAC. Point EE lies on AC‾\overline{AC}, and BE‾\overline{BE} bisects ∠ABC\angle ABC. The bisectors intersect at FF. What is the ratio AF:FDAF : FD?

Angle bisector ratio (its total is revealed at the end of the example)

DEF687ABC

The two halves of angle A are equal: 29° and 29°.

6 ÷ 3 = 2 and 8 ÷ 4 = 2
AF : FD = (6 + 8) : 7 = ? (hidden until the end of the example)

AB = 6, BD = 3, AC = 8, CD = 4: AB/BD = 2 and AC/CD = 2. The bisectors from A and B meet at F: AF : FD = (AB + AC) : BC = ?

  1. 1

    Use the Angle Bisector Theorem on ADAD: BD:DC=AB:AC=6:8BD : DC = AB : AC = 6 : 8, and BD+DC=7BD + DC = 7. What is BDBD?

    Predict first:
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Example 32.5

AIME

In triangle ABCABC, AB=20AB = 20 and AC=11AC = 11. The angle bisector of ∠A\angle A intersects BCBC at point DD, and point MM is the midpoint of ADAD. Let PP be the point of the intersection of ACAC and BMBM. The ratio of CPCP to PAPA can be expressed in the form mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm + n.

  1. 1

    First, the Angle Bisector Theorem. What is BD:DCBD : DC?

    Predict first:
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Contest cornerreal contest problems

Problem 32.2.7

AMC 10

Triangle ABCABC with AB=50AB = 50 and AC=10AC = 10 has area 120120. Let DD be the midpoint of AB‾\overline{AB}, and let EE be the midpoint of AC‾\overline{AC}. The angle bisector of ∠BAC\angle BAC intersects DE‾\overline{DE} and BC‾\overline{BC} at FF and GG, respectively. What is the area of quadrilateral FDBGFDBG?

Problem 32.2.11

AMC 10A

Let △ABC\triangle ABC be a scalene triangle. Point PP lies on BC‾\overline{BC} so that AP‾\overline{AP} bisects ∠BAC\angle BAC. The line through BB perpendicular to AP‾\overline{AP} intersects the line through AA parallel to BC‾\overline{BC} at point DD. Suppose BP=2BP = 2 and PC=3PC = 3. What is ADAD?

Click a choice or type its letter (A–E).

Problem 32.2.20(additional)

AIME

Trapezoid ABCDABCD has sides AB=92AB = 92, BC=50BC = 50, CD=19CD = 19, and AD=70AD = 70, with ABAB parallel to CDCD. A circle with center PP on ABAB is drawn tangent to BCBC and ADAD. Given that AP=mnAP = \frac{m}{n}, where mm and nn are relatively prime positive integers, find m+nm + n.

Problem 32.2.32(additional)

AMC 12Challenge

Let ABCABC be a triangle where MM is the midpoint of AC‾\overline{AC}, and CN‾\overline{CN} is the angle bisector of ∠ACB\angle ACB with NN on AB‾\overline{AB}. Let XX be the intersection of the median BM‾\overline{BM} and the bisector CN‾\overline{CN}. In addition △BXN\triangle BXN is equilateral with AC=2AC = 2. What is BX2BX^2?

Checkpoint for section 32.1

Get 2 of 3 right to master this section. One try each, no hints.
1

The altitude to the hypotenuse of a 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle is divided into two segments of lengths x<yx < y by the median to the shortest side of the triangle. What is the ratio xx+y\frac{x}{x + y}?

2

In △ABC\triangle ABC, AB=10AB = 10, AC=15AC = 15 and BC=10BC = 10. The bisector of ∠A\angle A meets BCBC at DD. What is BDBD?

3

In △ABC\triangle ABC, the bisector of ∠A\angle A meets BCBC at DD with BD=3BD = 3 and DC=5DC = 5. If AB=6AB = 6, what is ACAC?

32.2

Practice Problems

In this section you will be able to mix similar triangles, the right-triangle altitude and the Angle Bisector Theorem on harder contest problems.

Time to mix it up. In each problem, first find the pair of similar triangles. The algebra after that is short. Every problem has three hints if you get stuck.

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 32.2.3

AMC 10/12

Trapezoid ABCDABCD has AB‾∥CD‾\overline{AB} \parallel \overline{CD}, BC=CD=43BC = CD = 43, and AD‾⊥BD‾\overline{AD} \perp \overline{BD}. Let OO be the intersection of the diagonals AC‾\overline{AC} and BD‾\overline{BD}, and let PP be the midpoint of BD‾\overline{BD}. Given that OP=11OP = 11, the length ADAD can be written in the form mnm\sqrt{n}, where mm and nn are positive integers and nn is not divisible by the square of any prime. What is m+nm + n?

Problem 32.2.5

AMC 10

In quadrilateral ABCDABCD, ∠BAD≅∠ADC\angle BAD \cong \angle ADC and ∠ABD≅∠BCD\angle ABD \cong \angle BCD, AB=8AB = 8, BD=10BD = 10, and BC=6BC = 6. The length CDCD may be written in the form mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm + n.

Problem 32.2.9

AMC 12

Triangle ABCABC has AB=13AB = 13, BC=14BC = 14 and AC=15AC = 15. Let PP be the point on AC‾\overline{AC} such that PC=10PC = 10. There are exactly two points DD and EE on line BPBP such that quadrilaterals ABCDABCD and ABCEABCE are trapezoids. What is the distance DEDE?

Problem 32.2.10

AMC 10

This problem also appears in the book as 32.2.7.

Triangle ABCABC with AB=50AB = 50 and AC=10AC = 10 has area 120120. Let DD be the midpoint of AB‾\overline{AB}, and let EE be the midpoint of AC‾\overline{AC}. The angle bisector of ∠BAC\angle BAC intersects DE‾\overline{DE} and BC‾\overline{BC} at FF and GG, respectively. What is the area of quadrilateral FDBGFDBG?

Problem 32.2.14(additional)

AMC 10

Rectangle ABCDABCD has AB=6AB = 6 and BC=3BC = 3. Point MM is chosen on side AB‾\overline{AB} so that ∠AMD=∠CMD\angle AMD = \angle CMD. What is the degree measure of ∠AMD\angle AMD?

Problem 32.2.15(additional)

AIME

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region CC to the area of shaded region BB is 11/511/5. Find the ratio of shaded region DD to the area of shaded region AA.

Problem 32.2.16(additional)

AIME

Let PP be an interior point of triangle ABCABC and extend lines from the vertices through PP to the opposite sides. Let aa, bb, cc, and dd denote the lengths of the segments indicated in the figure. Find the product abcabc if a+b+c=43a + b + c = 43 and d=3d = 3.

Problem 32.2.17(additional)

AMC 12

In triangle ABCABC, AB=13AB = 13, BC=14BC = 14, and CA=15CA = 15. Distinct points DD, EE, and FF lie on segments BC‾\overline{BC}, CA‾\overline{CA}, and DE‾\overline{DE}, respectively, such that AD‾⊥BC‾\overline{AD} \perp \overline{BC}, DE‾⊥AC‾\overline{DE} \perp \overline{AC}, and AF‾⊥BF‾\overline{AF} \perp \overline{BF}. The length of segment DF‾\overline{DF} can be written as mn\frac{m}{n}, where mm and nn are relatively prime positive integers. What is m+nm + n?

Problem 32.2.18(additional)

AMC 10B

The altitude to the hypotenuse of a 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle is divided into two segments of lengths x<yx < y by the median to the shortest side of the triangle. What is the ratio xx+y\frac{x}{x + y}?

Problem 32.2.19(additional)

AMC 10

A square with side length xx is inscribed in a right triangle with sides of length 33, 44, and 55 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length yy is inscribed in another right triangle with sides of length 33, 44, and 55 so that one side of the square lies on the hypotenuse of the triangle. What is xy\frac{x}{y}?

Problem 32.2.21(additional)

AIME

In trapezoid ABCDABCD, leg BC‾\overline{BC} is perpendicular to bases AB‾\overline{AB} and CD‾\overline{CD}, and diagonals AC‾\overline{AC} and BD‾\overline{BD} are perpendicular. Given that AB=11AB = \sqrt{11} and AD=1001AD = \sqrt{1001}, find BC2BC^2.

Problem 32.2.22(additional)

AIME

Rhombus PQRSPQRS is inscribed in rectangle ABCDABCD so that vertices PP, QQ, RR, and SS are interior points on sides AB‾\overline{AB}, BC‾\overline{BC}, CD‾\overline{CD}, and DA‾\overline{DA}, respectively. It is given that PB=15PB = 15, BQ=20BQ = 20, PR=30PR = 30, and QS=40QS = 40. Let mn\frac{m}{n}, in lowest terms, denote the perimeter of ABCDABCD. Find m+nm + n.

Problem 32.2.23(additional)

AMC 10

In the right triangle △ACE\triangle ACE, we have AC=12AC = 12, CE=16CE = 16, and EA=20EA = 20. Points BB, DD, and FF are located on AC‾\overline{AC}, CE‾\overline{CE}, and EA‾\overline{EA}, respectively, so that AB=3AB = 3, CD=4CD = 4, and EF=5EF = 5. What is the ratio of the area of △DBF\triangle DBF to that of △ACE\triangle ACE?

Problem 32.2.25(additional)

AMC 10A

A silo (a right circular cylinder) with diameter 2020 meters stands in a field. MacDonald is located 2020 meters west and 1515 meters south of the center of the silo. McGregor is located 2020 meters east and g>0g > 0 meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of gg can be written as ab−cd\frac{a\sqrt{b} - c}{d}, where aa, bb, cc, dd are positive integers, bb is not divisible by the square of any prime, and dd is relatively prime to gcd⁡(a,c)\gcd(a, c). What is a+b+c+da + b + c + d?

Problem 32.2.26(additional)

AMC 10

In rectangle ABCDABCD, AB=1AB = 1, BC=2BC = 2, and points EE, FF, and GG are midpoints of BC‾\overline{BC}, CD‾\overline{CD}, and AD‾\overline{AD}, respectively. Point HH is the midpoint of GE‾\overline{GE}. What is the area of the shaded region?

Problem 32.2.28(additional)

AMC 10

Rectangle ABCDABCD has AB=4AB = 4 and BC=3BC = 3. Segment EF‾\overline{EF} is constructed through BB so that EF‾\overline{EF} is perpendicular to DB‾\overline{DB}, and AA and CC lie on DE‾\overline{DE} and DF‾\overline{DF}, respectively. What is EFEF?

Problem 32.2.30(additional)

AMC 10

In rectangle ABCDABCD, we have AB=8AB = 8, BC=9BC = 9, HH is on BC‾\overline{BC} with BH=6BH = 6, EE is on ADAD with DE=4DE = 4, line ECEC intersects line AHAH at GG, and FF is on line ADAD with GF‾⊥AF‾\overline{GF} \perp \overline{AF}. Find the length GFGF.

Problem 32.2.31(additional)

AIMEChallenge

Square ABCDABCD has sides of length 11. Points EE and FF are on BC‾\overline{BC} and CD‾\overline{CD}, respectively, so that △AEF\triangle AEF is equilateral. A square with vertex BB has sides that are parallel to those of ABCDABCD and a vertex on AE‾\overline{AE}. The length of a side of this smaller square is a−bc\frac{a - \sqrt{b}}{c}, where aa, bb, and cc are positive integers and bb is not divisible by the square of any prime. Find a+b+ca + b + c.