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Transcript
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:
𝐴𝐡 = 𝐷𝐢
Prove:
β–³ 𝐴𝐡𝐢 β‰… β–³ 𝐷𝐢𝐡
Given:
𝐴𝐡 = 𝑅𝑆
Μ…Μ…Μ…Μ…
𝐴𝐡 βˆ₯ Μ…Μ…Μ…Μ…
𝑅𝑆
Prove:
β–³ 𝐴𝐡𝐢 β‰… β–³ 𝑅𝑆𝑇
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 𝐴𝐡𝐢. 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 β–³ 𝐴𝐡𝐢, with 𝐴𝐡 = 𝐴𝐢
Prove:
π‘šβˆ π΅ = π‘šβˆ πΆ
βƒ—βƒ—βƒ—βƒ—βƒ— of ∠𝐴, where 𝐷 is the intersection of the
Construction: Draw the angle bisector 𝐴𝐷
Μ…Μ…Μ…Μ…
bisector and 𝐡𝐢 . We need to show that rigid motions maps point 𝐡 to point 𝐢 and
point 𝐢 to point 𝐡.
Let π‘Ÿ be the reflection through ⃑⃗⃗⃗⃗
𝐴𝐷 . Through the reflection, we want to demonstrate two
βƒ—βƒ—βƒ—βƒ—βƒ— , and (2) 𝐴𝐡 = 𝐴𝐢.
pieces of information that map 𝐡 to point 𝐢 and vice versa: (1) βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐡 maps to 𝐴𝐢
Since 𝐴 is on the line of reflection, ⃑⃗⃗⃗⃗
𝐴𝐷 , π‘Ÿ(𝐴) = 𝐴. Reflections preserve angle measures, so the measure of the reflected
βƒ—βƒ—βƒ—βƒ—βƒ— ) = 𝐴𝐢
βƒ—βƒ—βƒ—βƒ—βƒ— . Reflections also preserve lengths of segments;
angle π‘Ÿ(∠𝐡𝐴𝐷) equals the measure of ∠𝐢𝐴𝐷; therefore, π‘Ÿ(𝐴𝐡
Μ…Μ…Μ…Μ…
Μ…Μ…Μ…Μ…
therefore, the reflection of 𝐴𝐡 still has the same length as 𝐴𝐡 . By hypothesis, 𝐴𝐡 = 𝐴𝐢, so the length of the reflection
is also equal to 𝐴𝐢. Then π‘Ÿ(𝐡) = 𝐢. Using similar reasoning, we can show that π‘Ÿ(𝐢) = 𝐡.
βƒ—βƒ—βƒ—βƒ—βƒ— ) = 𝐢𝐴
βƒ—βƒ—βƒ—βƒ—βƒ— and π‘Ÿ(𝐡𝐢
βƒ—βƒ—βƒ—βƒ—βƒ— ) = 𝐢𝐡
βƒ—βƒ—βƒ—βƒ—βƒ— . Again, since reflections preserve angle measures, the
Reflections map rays to rays, so π‘Ÿ(𝐡𝐴
measure of π‘Ÿ(∠𝐴𝐡𝐢) is equal to the measure of ∠𝐴𝐢𝐡.
We conclude that π‘šβˆ π΅ = π‘šβˆ πΆ. Equivalently, we can state that ∠𝐡 β‰… ∠𝐢. In proofs, we can state that β€œbase angles of
an isosceles triangle are equal in measure” or that β€œbase angles of an isosceles triangle are congruent.”
Prove Base Angles of an Isosceles are Congruent: SAS
Given:
Isosceles β–³ 𝐴𝐡𝐢, with 𝐴𝐡 = 𝐴𝐢
Prove:
∠𝐡 β‰… ∠𝐢
Construction: Draw the angle bisector βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐷 of ∠𝐴, where 𝐷 is the intersection of the
bisector and Μ…Μ…Μ…Μ…Μ…
𝐡𝐢. 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 𝐾𝐿
Μ…Μ…Μ…Μ…
𝐽𝐾 = 𝐽𝐿; 𝐽𝑅
Μ…Μ…Μ…
Μ…Μ…Μ…Μ…
𝐽𝑅 βŠ₯ 𝐾𝐿
Prove:
𝐴𝐡 = 𝐴𝐢, 𝑋𝐡 = 𝑋𝐢
Μ…Μ…Μ…Μ…
𝐴𝑋 bisects ∠𝐡𝐴𝐢
Given:
𝐽𝑋 = π½π‘Œ, 𝐾𝑋 = πΏπ‘Œ
Prove:
β–³ 𝐽𝐾𝐿 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:
β–³ 𝐴𝐡𝐢, with π‘šβˆ πΆπ΅π΄ = π‘šβˆ π΅πΆπ΄
Prove:
𝐡𝐴 = 𝐢𝐴
(Converse of base angles of isosceles triangle)
Hint: Use a transformation.
5.
Given:
Μ…Μ…Μ…Μ… βˆ₯ π‘‹π‘Œ
Μ…Μ…Μ…Μ…Μ…is the angle bisector of βˆ π΅π‘Œπ΄, and 𝐡𝐢
Μ…Μ…Μ…Μ…
β–³ 𝐴𝐡𝐢, with π‘‹π‘Œ
Prove:
π‘Œπ΅ = π‘ŒπΆ
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:
𝐴𝐡 = 𝐡𝐢, 𝐴𝐷 = 𝐷𝐢
Prove:
β–³ 𝐴𝐷𝐡 and β–³ 𝐢𝐷𝐡 are right triangles
Given:
Μ…Μ…Μ…Μ…
𝐴𝐢 = 𝐴𝐸 and Μ…Μ…Μ…Μ…
𝐡𝐹 ‖𝐢𝐸
Prove:
𝐴𝐡 = 𝐴𝐹
Μ…Μ…Μ…Μ… β‰… Μ…Μ…Μ…Μ…
In the diagram, β–³ 𝐴𝐡𝐢 is isosceles with 𝐴𝐢
𝐴𝐡 . In your own words, describe
how transformations and the properties of rigid motions can be used to show
that ∠𝐢 β‰… ∠𝐡.
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.