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Lesson 23: Base Angles of Isosceles Triangles NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 M1 GEOMETRY Lesson 23: Base Angles of Isosceles Triangles Student Outcomes . Students examine two different proof techniques via a familiar theorem. Students complete proofs involving properties of an isosceles triangle. Lesson Notes In Lesson 23, students study two proofs to demonstrate why the base angles of isosceles triangles are congruent. The first proof uses transformations, while the second uses the recently acquired understanding of the SAS triangle congruency. The demonstration of both proofs highlight the utility of the SAS criteria. Encourage students to articulate why the SAS criteria is so useful. The goal of this lesson is to compare two different proof techniques by investigating a familiar theorem. Be careful not to suggest that proving an established fact about isosceles triangles is somehow the significant aspect of this lesson. However, if necessary, you can help motivate the lesson by appealing to history. Point out that Euclid used SAS and that the first application he made of it was in proving that base angles of an isosceles triangle are congruent. This is a famous part of the Elements. Today, proofs using SAS and proofs using rigid motions are valued equally. Classwork Opening Exercise (6 minutes) Opening Exercise Describe the additional piece of information needed for each pair of triangles to satisfy the SAS triangle congruence criteria. a. Given: 푨푩 = 푫푪 푨푩̅̅̅̅ ⊥ 푩푪̅̅̅̅, 푫푪̅̅̅̅ ⊥ 푪푩̅̅̅̅, or 풎∠푩 = 풎∠푪 Prove: △ 푨푩푪 ≅ △ 푫푪푩 b. Given: 푨푩 = 푹푺 푨푩̅̅̅̅ ∥ 푹푺̅̅̅̅ 푪푩 = 푻푺, or 푪푻 = 푩푺 Prove: △ 푨푩푪 ≅ △ 푹푺푻 Lesson 23: Base Angles of Isosceles Triangles 200 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 M1 GEOMETRY Exploratory Challenge (25 minutes) Exploratory Challenge Today we examine a geometry fact that we already accept to be true. We are going to prove this known fact in two ways: (1) by using transformations and (2) by using SAS triangle congruence criteria. Here is isosceles triangle 푨푩푪. We accept that an isosceles triangle, which has (at least) two congruent sides, also has congruent base angles. Label the congruent angles in the figure. Now we prove that the base angles of an isosceles triangle are always congruent. Prove Base Angles of an Isosceles are Congruent: Transformations While simpler proofs do exist, this version is included to reinforce the idea of congruence as defined by rigid motions. Allow 15 minutes for the first proof. Prove Base Angles of an Isosceles are Congruent: Transformations Given: Isosceles △ 푨푩푪, with 푨푩 = 푨푪 Prove: 풎∠푩 = 풎∠푪 Construction: Draw the angle bisector 푨푫⃗⃗⃗⃗⃗⃗ of ∠푨, where 푫 is the intersection of the bisector and 푩푪̅̅̅̅. We need to show that rigid motions maps point 푩 to point 푪 and point 푪 to point 푩. Let 풓 be the reflection through 푨푫⃡⃗⃗⃗⃗ . Through the reflection, we want to demonstrate two pieces of information that map 푩 to point 푪 and vice versa: (1) 푨푩⃗⃗⃗⃗⃗⃗ maps to 푨푪⃗⃗⃗⃗⃗ , and (2) 푨푩 = 푨푪. Since 푨 is on the line of reflection, 푨푫⃡⃗⃗⃗⃗ , 풓(푨) = 푨. Reflections preserve angle measures, so the measure of the reflected angle 풓(∠푩푨푫) equals the measure of ∠푪푨푫; therefore, 풓(푨푩⃗⃗⃗⃗⃗⃗ ) = 푨푪⃗⃗⃗⃗⃗ . Reflections also preserve lengths of segments; therefore, the reflection of 푨푩̅̅̅̅ still has the same length as 푨푩̅̅̅̅. By hypothesis, 푨푩 = 푨푪, so the length of the reflection is also equal to 푨푪. Then 풓(푩) = 푪. Using similar reasoning, we can show that 풓(푪) = 푩. Reflections map rays to rays, so 풓(푩푨⃗⃗⃗⃗⃗⃗ ) = 푪푨⃗⃗⃗⃗⃗ and 풓(푩푪⃗⃗⃗⃗⃗⃗ ) = 푪푩⃗⃗⃗⃗⃗⃗ . Again, since reflections preserve angle measures, the measure of 풓(∠푨푩푪) is equal to the measure of ∠푨푪푩. We conclude that 풎∠푩 = 풎∠푪. Equivalently, we can state that ∠푩 ≅ ∠푪. In proofs, we can state that “base angles of an isosceles triangle are equal in measure” or that “base angles of an isosceles triangle are congruent.” Lesson 23: Base Angles of Isosceles Triangles 201 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 M1 GEOMETRY Prove Base Angles of an Isosceles are Congruent: SAS Allow 10 minutes for the second proof. Prove Base Angles of an Isosceles are Congruent: SAS Given: Isosceles △ 푨푩푪, with 푨푩 = 푨푪 Prove: ∠푩 ≅ ∠푪 Construction: Draw the angle bisector 푨푫⃗⃗⃗⃗⃗⃗ of ∠푨, where 푫 is the intersection of the bisector and 푩푪̅̅̅̅. We are going to use this auxiliary line towards our SAS criteria. Allow students five minutes to attempt the proof on their own. 푨푩 = 푨푪 Given 푨푫 = 푨푫 Reflexive property 풎∠푩푨푫 = 풎∠푪푨푫 Definition of an angle bisector △ 푨푩푫 ≅ △ 푨푪푫 SAS ∠푩 ≅ ∠푪 Corresponding angles of congruent triangles are congruent. Exercises (10 minutes) Note that Exercise 4 asks students to use transformations in the proof. This is intended to reinforce the notion that congruence is defined in terms of rigid motions. Exercises 1. Given: 푱푲 = 푱푳; 푱푹̅̅̅̅ bisects 푲푳̅̅̅̅ Prove: 푱푹̅̅̅̅ ⊥ 푲푳̅̅̅̅ 푱푲 = 푱푳 Given ∠푲 ≅ ∠푳 Base angles of an isosceles triangle are congruent. 푲푹 = 푹푳 Definition of a segment bisector ∴ △ 푱푹푲 ≅ △ 푱푹푳 SAS ∠푱푹푲 ≅ ∠푱푹푳 Corresponding angles of congruent triangles are congruent. 풎∠푱푹푲 + 풎∠푱푹푳 = ퟏퟖퟎ° Linear pairs form supplementary angles. ퟐ(풎∠푱푹푲) = ퟏퟖퟎ° Substitution property of equality 풎∠푱푹푲 = ퟗퟎ° Division property of equality ∴ ̅푱̅̅푹̅ ⊥ 푲푳̅̅̅̅ Definition of perpendicular lines Lesson 23: Base Angles of Isosceles Triangles 202 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 M1 GEOMETRY 2. Given: 푨푩 = 푨푪, 푿푩 = 푿푪 Prove: 푨푿̅̅̅̅ bisects ∠푩푨푪 푨푩 = 푨푪 Given 풎∠푨푩푪 = 풎∠푨푪푩 Base angles of an isosceles triangle are equal in measure. 푿푩 = 푿푪 Given 풎∠푿푩푪 = 풎∠푿푪푩 Base angles of an isosceles triangle are equal in measure. 풎∠푨푩푪 = 풎∠푨푩푿 + 풎∠푿푩푪, 풎∠푨푪푩 = 풎∠푨푪푿 + 풎∠푿푪푩 Angle addition postulate 풎∠푨푩푿 = 풎∠푨푩푪 − 풎∠푿푩푪, 풎∠푨푪푿 = 풎∠푨푪푩 − 풎∠푿푪푩 Subtraction property of equality 풎∠푨푩푿 = 풎∠푨푪푿 Substitution property of equality ∴ △ 푨푩푿 ≅ △ 푨푪푿 SAS ∠푩푨푿 ≅ ∠푪푨푿 Corresponding parts of congruent triangles are congruent. ∴ 푨푿̅̅̅̅ bisects ∠푩푨푪. Definition of an angle bisector 3. Given: 푱푿 = 푱풀, 푲푿 = 푳풀 Prove: △ 푱푲푳 is isosceles 푱푿 = 푱풀 Given 푲푿 = 푳풀 Given 풎∠ퟏ = 풎∠ퟐ Base angles of an isosceles triangle are equal in measure. 풎∠ퟏ + 풎∠ퟑ = ퟏퟖퟎ° Linear pairs form supplementary angles. 풎∠ퟐ + 풎∠ퟒ = ퟏퟖퟎ° Linear pairs form supplementary angles. 풎∠ퟑ = ퟏퟖퟎ° − 풎∠ퟏ Subtraction property of equality 풎∠ퟒ = ퟏퟖퟎ° − 풎∠ퟑ Subtraction property of equality 풎∠ퟑ = 풎∠ퟒ Substitution property of equality ∴ △ 푱푲푿 ≅△ 푱풀푳 SAS ̅푱푲̅̅̅ ≅ 푱푳̅̅̅ Corresponding parts of congruent triangles are congruent. ∴ △ 푱푲푳 is isosceles. Definition of an isosceles triangle Lesson 23: Base Angles of Isosceles Triangles 203 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 M1 GEOMETRY 4. Given: △ 푨푩푪, with 풎∠푪푩푨 = 풎∠푩푪푨 Prove: 푩푨 = 푪푨 (Converse of base angles of isosceles triangle) Hint: Use a transformation. PROOF: We can prove that 푨푩 = 푨푪 by using rigid motions. Construct the perpendicular bisector 풍 to 푩푪̅̅̅̅. Note that we do not know whether the point 푨 is on 풍. If we did, we would know immediately that 푨푩 = 푨푪, since all points on the perpendicular bisector are equidistant from 푩 and 푪. We need to show that 푨 is on 풍, or equivalently, that the reflection across 풍 takes 푨 to 푨. Let 풓풍 be the transformation that reflects the △ 푨푩푪 across 풍. Since 풍 is the perpendicular bisector, 풓풍(푩) = 푪 and ′ 풓풍(푪) = 푩. We do not know where the transformation takes 푨; for now, let us say that 풓풍(푨) = 푨 . Since ∠푪푩푨 ≅ ∠푩푪푨 and rigid motions preserve angle measures, after applying 풓풍 to ∠푩푪푨, we get that ∠푪푩푨 ≅ ∠푪푩푨′. ∠푪푩푨 and ∠푪푩푨′ share a ray 푩푪, are of equal measure, and 푨 and 푨′ are both in the same half- plane with respect to line 푩푪. Hence, 푩푨⃗⃗⃗⃗⃗⃗ and ⃗푩푨⃗⃗⃗⃗⃗⃗ ′ are the same ray. In particular, 푨′ is a point somewhere on 푩푨⃗⃗⃗⃗⃗⃗ . ′ Likewise, applying 풓풍 to ∠푪푩푨 gives ∠푩푪푨 ≅ ∠푩푪푨 , and for the same reasons in the previous paragraph, 푨′ must also be a point somewhere on 푪푨⃗⃗⃗⃗⃗ . Therefore, 푨′ is common to both 푩푨⃗⃗⃗⃗⃗⃗ and 푪푨⃗⃗⃗⃗⃗ . The only point common to both 푩푨⃗⃗⃗⃗⃗⃗ and 푪푨⃗⃗⃗⃗⃗ is point 푨; thus, 푨 and 푨′ must be the same point, i.e., 푨 = 푨′. ′ Hence, the reflection takes 푨 to 푨, which means 푨 is on the line 풍 and 풓풍(푩푨) = 푪푨 = 푪푨, or 푩푨 = 푪푨. 5. Given: △ 푨푩푪, with 푿풀̅̅̅̅̅ is the angle bisector of ∠푩풀푨, and 푩푪̅̅̅̅ ∥ 푿풀̅̅̅̅ Prove: 풀푩 = 풀푪 푩푪̅̅̅̅ ∥ 푿풀̅̅̅̅ Given 풎∠푿풀푩 = 풎∠푪푩풀 When two parallel lines are cut by a transversal, the alternate interior angles are equal. 풎∠푿풀푨 = 풎∠푩푪풀 When two parallel lines are cut by a transversal, the corresponding angles are equal.
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