Math 135 Similar Triangles Definition of Similar Triangles ∆ABC is similar to ∆DEF (written ∆ABC ~∆DEF ) under the correspondence A ↔ D,B ↔ E,C ↔ F if and only if: 1) All three pairs of corresponding angles are congruent. 2) All pairs of corresponding sides are proportional. D A B C E F AB BC AC ∆ABC ~ ∆DEF iff ∠A ≅ ∠D : ∠B ≅ ∠E : ∠C ≅ ∠F and = = DE EF DF AAA Similarity Postulate Two triangles are similar if and only if three angles of one triangle are congruent, respectively, to three angles of the other triangle. ∆ABC ~ ∆DEF iff ∠A ≅ ∠D : ∠B ≅ ∠E : ∠C ≅ ∠F AA Similarity Theorem Two triangles are similar if two angles of one triangle are congruent, respectively to two angles of the other triangle. In ∆ABC if ∠A ≅ ∠D and ∠B ≅ ∠E, then ∆ABC ~∆DEF Proof: Given ∠A ≅ ∠D and ∠B ≅ ∠E , show that ∆ABC ~∆DEF D A B C E F Let ∠A ≅ ∠D and ∠B ≅ ∠E ⇒ m∠A = m∠D and m∠B = m∠E Since the some of the interior angles is 180 degrees, m∠A + m∠B + m∠C = 180 ° and m∠D + m∠E + m∠F = 180 ° . Use the substitution principle on the second equation with the values m∠A = m∠D and m∠B = m∠E to get the following equation. m∠A + m∠B + m∠F = 180 °. Subtracting the two equations gives: m∠A + m∠B + m∠C = 180 ° m∠A + m∠B + m∠C = 180 ° − ()m∠A + m∠B + m∠F = 180 ° ⇒ − m∠A − m∠B − m∠F = −180 ° m∠C − m∠F = 0° m∠C = m∠F ⇒ ∠C ≅ ∠F Therefore ∆ABC ~∆DEF by AAA Similarity Postulate Example 1 In ∆ABC , DE is parallel to AC . If AB = 10, CE = ,4 AC = ,8 BD = 6 and BE = x , find DE and BC . B x 6 10 D E 4 A 8 C Solution: 6 x = 10 x + 4 Taking the product of the means and the extremes, get the following: 10 x = 6(x + 4) 10 x = 6x + 24 10 x − 6x = 6x − 6x + 24 4x = 24 4x 24 = 4 4 ⇒ x = 6 ⇒ BE = 6 Find BC 6 BC = 10 8 10 BC = 48 BC = 8.4 Corollary 6.5 Two right triangles are similar if an acute angle of one triangle is congruent to an acute angle of the other triangle. D A B C E F Theorem 6.6 (SAS Similarity Theorem) Two triangles are similar if two sides are proportional, respectively, to two sides of another triangle and the angles included between the sides are congruent. Corollary 6.7 (LL Similarity) Two right triangles are similar if the legs of one triangle are proportional to respectively to the legs of the other triangle. D A B C E F Theorem 6.8 (SSS Similarity Theorem) Two triangles are similar if three sides of one triangle are proportional to three sides of the other triangle. D A B C E F Example 2 Given ∆CAB ~∆CED , find the missing measures. A 6 12 10 D C ? B 18 ? E BC 10 = 12 18 18 ⋅ BC = 12 ⋅10 18 ⋅ BC = 120 120 2 ⇒ BC = = 6 18 3 CE 6 = 18 12 12 ⋅ CE = 18 ⋅ 6 12 ⋅ CE = 108 108 ⇒ CE = = 9 12 Example 3 Given ∆CAB ~∆CED , find the missing measures. A ? C 8 10 5 D ? B ? E Solution: Let AC = x 10 10 + x = 5 8 5()10 + x = 10 ⋅8 50 + 5x = 80 50 − 50 + 5x = 80 − 50 5x = 30 30 x = = 6 ⇒ AE = 16 5 Use the Pythagorean Theorem to find BE and DB (BE )2 + 52 = 10 2 (DE )2 + 82 = 16 2 ()BE 2 + 25 = 100 ()DE 2 + 64 = 256 ()BE 2 = 75 ()DE 2 = 192 ()BE 2 = 75 ()DE 2 = 192 BE = 5 3 DE = 36 ⋅ 3 = 6 3 Therefore, BE = 5 3 and DB = 6 3 − 5 3 = 3 Example 4 Compute the ratios requested in the following pairs of similar figures. Find the ratios of base: base, height: height, and area; area for the following pair of similar rectangles. 15" 5" 2" 6" Bases: 2:6 or 1:3 Heights: 5:15 or 1:3 Areas: 10:90 or 1:9 (See below) Area of smaller rectangle A1 = 2 ⋅5 = 10 Area of larger rectangle A1 = 6 ⋅15 = 90 Example 5 Show that the area of the larger triangle shown in red is equal to the sum of the smaller two triangles C B 5 cm 3 cm E 4 cm Triangle with Hypotenuse = 3 Find the length of the leg using Pythagorean Theorem a 2 + b 2 = c 2 x 2 + x 2 = 32 2x 2 = 9 9 x 2 = 2 9 3 x = = 2 2 3 Length of leg = 2 1 1 3 3 9 Area 1 = bh = ⋅ ⋅ = 2 2 2 2 4 Triangle with Hypotenuse = 4 Find the length of the leg using Pythagorean Theore a 2 + b 2 = c 2 x 2 + x 2 = 42 2x 2 = 16 16 x 2 = 2 x = 8 = 2 2 Length of leg = 2 2 1 1 1 Area = bh = ⋅ 2 2 ⋅ 2 2 = ⋅8 = 4 2 2 2 2 Triangle with Hypotenuse = 5 Find the length of the leg using Pythagorean Theore a 2 + b 2 = c 2 x 2 + x 2 = 52 2x 2 = 25 25 x 2 = 2 25 5 x = = 2 2 5 Length of leg = 2 1 1 5 5 15 Area 3 = bh = ⋅ ⋅ = 2 2 2 2 4 Check: Area 1 + Area 2 = Area 3 9 25 9 16 25 + 4 = ⇒ + = 4 4 4 4 4 Example 6 In the figure shown, ∠A ≅ ∠C , M and N are midpoints of the sides AB and BC and MNPQ is a rectangle . Show that ∆MQA ≅ ∆NPC . B M N A Q P C Statement Reason ∠A ≅ ∠C Given M and N are midpoints of the sides AB and BC CB ≅ AB ∆ABC is a isosceles triangle MQ ≅ NP Opposite sides of a rectangle are congruent. ∠AQM and ∠CPN are right angles The exterior angle of a rectangle are right angles ∠AMQ ≅ ∠CNP If two angles of one triangle are congruent to two angle of another triangle, then the remaining angles are congruent. ∆MQA ≅ ∆NPC ASA Find the area of MNPQ where MN = 14 m and BC = 22 m B 11 14 m M N 22 7 11 j Q P A C c 2 = a 2 + b 2 11 2 = 7 2 + b 2 121 = 49 + b 2 73 = b 2 ⇒ b = 72 = 6 2 NP = 6 2 Area = 14 • 6 2 = 84 2 ≈ 118 79. m 2 Example 7 Explain how the figure can be used to find a B x A D C a 1 a x = x 1 x ⋅ x = a ⋅1 x 2 = a x 2 = a x = a Example 8 In the figure shown A,B,C, and D are midpoints of the sides of parallelogram PQRS. If the area of ∆AQB = 5 in 2 , find the area of hexagon PABRCD. Q B R A C P D S The parallelogram can be divided into 8 congruent triangles and the area of the hexagon can be found by adding up six of these congruent triangles. See below: Q B R 5 5 5 T A C 5 5 5 P D S .