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Lesson 23 NYS COMMON CORE MATHEMATICS CURRICULUM M1 GEOMETRY Lesson 23: Base Angles of Isosceles Triangles Classwork Opening Exercise Describe the additional piece of information needed for each pair of triangles to satisfy the SAS triangle congruence criteria. a. b. Given: π΄π΅ = π·πΆ Prove: β³ π΄π΅πΆ β β³ π·πΆπ΅ Given: π΄π΅ = π π Μ Μ Μ Μ π΄π΅ β₯ Μ Μ Μ Μ π π Prove: β³ π΄π΅πΆ β β³ π ππ 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 π΄π΅πΆ. 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. Lesson 23: Base Angles of Isosceles Triangles This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 S.125 This work is licensed under a 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: Transformations Given: Isosceles β³ π΄π΅πΆ, with π΄π΅ = π΄πΆ Prove: πβ π΅ = πβ πΆ βββββ of β π΄, where π· is the intersection of the Construction: Draw the angle bisector π΄π· Μ Μ Μ Μ bisector and π΅πΆ . We need to show that rigid motions maps point π΅ to point πΆ and point πΆ to point π΅. Let π be the reflection through β‘ββββ π΄π· . Through the reflection, we want to demonstrate two βββββ , and (2) π΄π΅ = π΄πΆ. pieces of information that map π΅ to point πΆ and vice versa: (1) βββββ π΄π΅ maps to π΄πΆ Since π΄ is on the line of reflection, β‘ββββ π΄π· , π(π΄) = π΄. Reflections preserve angle measures, so the measure of the reflected βββββ ) = π΄πΆ βββββ . Reflections also preserve lengths of segments; angle π(β π΅π΄π·) equals the measure of β πΆπ΄π·; therefore, π(π΄π΅ Μ Μ Μ Μ Μ Μ Μ Μ therefore, the reflection of π΄π΅ still has the same length as π΄π΅ . By hypothesis, π΄π΅ = π΄πΆ, so the length of the reflection is also equal to π΄πΆ. Then π(π΅) = πΆ. Using similar reasoning, we can show that π(πΆ) = π΅. βββββ ) = πΆπ΄ βββββ and π(π΅πΆ βββββ ) = πΆπ΅ βββββ . Again, since reflections preserve angle measures, the Reflections map rays to rays, so π(π΅π΄ measure of π(β π΄π΅πΆ) is equal to the measure of β π΄πΆπ΅. We conclude that πβ π΅ = πβ πΆ. Equivalently, we can state that β π΅ β β πΆ. In proofs, we can state that βbase angles of an isosceles triangle are equal in measureβ or that βbase angles of an isosceles triangle are congruent.β Prove Base Angles of an Isosceles are Congruent: SAS Given: Isosceles β³ π΄π΅πΆ, with π΄π΅ = π΄πΆ Prove: β π΅ β β πΆ Construction: Draw the angle bisector βββββ π΄π· of β π΄, where π· is the intersection of the bisector and Μ Μ Μ Μ Μ π΅πΆ. We are going to use this auxiliary line towards our SAS criteria. Lesson 23: Base Angles of Isosceles Triangles This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 S.126 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 M1 GEOMETRY Exercises 1. Given: Prove: 2. 3. Μ Μ Μ bisects πΎπΏ Μ Μ Μ Μ π½πΎ = π½πΏ; π½π Μ Μ Μ Μ Μ Μ Μ π½π β₯ πΎπΏ Prove: π΄π΅ = π΄πΆ, ππ΅ = ππΆ Μ Μ Μ Μ π΄π bisects β π΅π΄πΆ Given: π½π = π½π, πΎπ = πΏπ Prove: β³ π½πΎπΏ is isosceles Given: Lesson 23: Base Angles of Isosceles Triangles This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 S.127 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 M1 GEOMETRY 4. Given: β³ π΄π΅πΆ, with πβ πΆπ΅π΄ = πβ π΅πΆπ΄ Prove: π΅π΄ = πΆπ΄ (Converse of base angles of isosceles triangle) Hint: Use a transformation. 5. Given: Μ Μ Μ Μ β₯ ππ Μ Μ Μ Μ Μ is the angle bisector of β π΅ππ΄, and π΅πΆ Μ Μ Μ Μ β³ π΄π΅πΆ, with ππ Prove: ππ΅ = ππΆ Lesson 23: Base Angles of Isosceles Triangles This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 S.128 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 23 M1 GEOMETRY Problem Set 1. 2. 3. Given: π΄π΅ = π΅πΆ, π΄π· = π·πΆ Prove: β³ π΄π·π΅ and β³ πΆπ·π΅ are right triangles Given: Μ Μ Μ Μ π΄πΆ = π΄πΈ and Μ Μ Μ Μ π΅πΉ βπΆπΈ Prove: π΄π΅ = π΄πΉ Μ Μ Μ Μ β Μ Μ Μ Μ In the diagram, β³ π΄π΅πΆ is isosceles with π΄πΆ π΄π΅ . In your own words, describe how transformations and the properties of rigid motions can be used to show that β πΆ β β π΅. Lesson 23: Base Angles of Isosceles Triangles This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M1-TE-1.3.0-07.2015 S.129 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.