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Transcript
Lesson 23
NYS COMMON CORE MATHEMATICS CURRICULUM
M1
GEOMETRY
Lesson 23: Base Angles of Isosceles Triangles
Student Outcomes
๏‚ง
Students examine two different proof techniques via a familiar theorem.
๏‚ง
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:
๐‘จ๐‘ฉ = ๐‘ซ๐‘ช
ฬ…ฬ…ฬ…ฬ…, ๐‘ซ๐‘ช
ฬ…ฬ…ฬ…ฬ… โŠฅ ๐‘ช๐‘ฉ
ฬ…ฬ…ฬ…ฬ…, or ๐’Žโˆ ๐‘ฉ = ๐’Žโˆ ๐‘ช
ฬ…ฬ…ฬ…ฬ… โŠฅ ๐‘ฉ๐‘ช
๐‘จ๐‘ฉ
b.
Prove:
โ–ณ ๐‘จ๐‘ฉ๐‘ช โ‰… โ–ณ ๐‘ซ๐‘ช๐‘ฉ
Given:
๐‘จ๐‘ฉ = ๐‘น๐‘บ
ฬ…ฬ…ฬ…ฬ…
ฬ…ฬ…ฬ…ฬ… โˆฅ ๐‘น๐‘บ
๐‘จ๐‘ฉ
๐‘ช๐‘ฉ = ๐‘ป๐‘บ, or ๐‘ช๐‘ป = ๐‘ฉ๐‘บ
Prove:
Lesson 23:
โ–ณ ๐‘จ๐‘ฉ๐‘ช โ‰… โ–ณ ๐‘น๐‘บ๐‘ป
Base Angles of Isosceles Triangles
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M1-TE-1.3.0-07.2015
200
This work is licensed under a
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 ๐‘จ๐‘ฉ๐‘ช. 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 โ–ณ ๐‘จ๐‘ฉ๐‘ช, with ๐‘จ๐‘ฉ = ๐‘จ๐‘ช
Prove: ๐’Žโˆ ๐‘ฉ = ๐’Žโˆ ๐‘ช
Construction: Draw the angle bisector โƒ—โƒ—โƒ—โƒ—โƒ—โƒ—
๐‘จ๐‘ซ of โˆ ๐‘จ, where ๐‘ซ is the intersection of the
ฬ…ฬ…ฬ…ฬ…. We need to show that rigid motions maps point ๐‘ฉ to point ๐‘ช and
bisector and ๐‘ฉ๐‘ช
point ๐‘ช to point ๐‘ฉ.
โƒกโƒ—โƒ—โƒ—โƒ— . Through the reflection, we want to demonstrate two pieces of information that map
Let ๐’“ be the reflection through ๐‘จ๐‘ซ
โƒ—โƒ—โƒ—โƒ—โƒ— , and (2) ๐‘จ๐‘ฉ = ๐‘จ๐‘ช.
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— maps to ๐‘จ๐‘ช
๐‘ฉ to point ๐‘ช and vice versa: (1) ๐‘จ๐‘ฉ
Since ๐‘จ is on the line of reflection, โƒกโƒ—โƒ—โƒ—โƒ—
๐‘จ๐‘ซ, ๐’“(๐‘จ) = ๐‘จ. Reflections preserve angle measures, so the measure of the reflected
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— ) = โƒ—โƒ—โƒ—โƒ—โƒ—
angle ๐’“(โˆ ๐‘ฉ๐‘จ๐‘ซ) equals the measure of โˆ ๐‘ช๐‘จ๐‘ซ; therefore, ๐’“(๐‘จ๐‘ฉ
๐‘จ๐‘ช. Reflections also preserve lengths of segments;
ฬ…ฬ…ฬ…ฬ… still has the same length as ๐‘จ๐‘ฉ
ฬ…ฬ…ฬ…ฬ…. By hypothesis, ๐‘จ๐‘ฉ = ๐‘จ๐‘ช, so the length of the reflection is
therefore, the reflection of ๐‘จ๐‘ฉ
also equal to ๐‘จ๐‘ช. Then ๐’“(๐‘ฉ) = ๐‘ช. Using similar reasoning, we can show that ๐’“(๐‘ช) = ๐‘ฉ.
โƒ—โƒ—โƒ—โƒ—โƒ— and ๐’“(๐‘ฉ๐‘ช
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— ) = ๐‘ช๐‘ฉ
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— . Again, since reflections preserve angle measures, the
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— ) = ๐‘ช๐‘จ
Reflections map rays to rays, so ๐’“(๐‘ฉ๐‘จ
measure of ๐’“(โˆ ๐‘จ๐‘ฉ๐‘ช) is equal to the measure of โˆ ๐‘จ๐‘ช๐‘ฉ.
We conclude that ๐’Žโˆ ๐‘ฉ = ๐’Žโˆ ๐‘ช. Equivalently, we can state that โˆ ๐‘ฉ โ‰… โˆ ๐‘ช. In proofs, we can state that โ€œbase angles of
an isosceles triangle are equal in measureโ€ or that โ€œbase angles of an isosceles triangle are congruent.โ€
Lesson 23:
Base Angles of Isosceles Triangles
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M1-TE-1.3.0-07.2015
201
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 23
NYS COMMON CORE MATHEMATICS CURRICULUM
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 โ–ณ ๐‘จ๐‘ฉ๐‘ช, with ๐‘จ๐‘ฉ = ๐‘จ๐‘ช
Prove: โˆ ๐‘ฉ โ‰… โˆ ๐‘ช
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— of โˆ ๐‘จ, where ๐‘ซ is the intersection of the
Construction: Draw the angle bisector ๐‘จ๐‘ซ
bisector and ฬ…ฬ…ฬ…ฬ…
๐‘ฉ๐‘ช. We are going to use this auxiliary line towards our SAS criteria.
Allow students five minutes to attempt the proof on their own.
๐‘จ๐‘ฉ = ๐‘จ๐‘ช
Given
๐‘จ๐‘ซ = ๐‘จ๐‘ซ
Reflexive property
๐’Žโˆ ๐‘ฉ๐‘จ๐‘ซ = ๐’Žโˆ ๐‘ช๐‘จ๐‘ซ
Definition of an angle bisector
โ–ณ ๐‘จ๐‘ฉ๐‘ซ โ‰… โ–ณ ๐‘จ๐‘ช๐‘ซ
SAS
โˆ ๐‘ฉ โ‰… โˆ ๐‘ช
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.
ฬ…ฬ…ฬ…ฬ… bisects ๐‘ฒ๐‘ณ
ฬ…ฬ…ฬ…ฬ…
๐‘ฑ๐‘ฒ = ๐‘ฑ๐‘ณ; ๐‘ฑ๐‘น
ฬ…ฬ…ฬ…ฬ… โŠฅ ๐‘ฒ๐‘ณ
ฬ…ฬ…ฬ…ฬ…
๐‘ฑ๐‘น
Given:
Prove:
๐‘ฑ๐‘ฒ = ๐‘ฑ๐‘ณ
Given
โˆ ๐‘ฒ โ‰… โˆ ๐‘ณ
Base angles of an isosceles triangle are congruent.
๐‘ฒ๐‘น = ๐‘น๐‘ณ
Definition of a segment bisector
โˆด โ–ณ ๐‘ฑ๐‘น๐‘ฒ โ‰… โ–ณ ๐‘ฑ๐‘น๐‘ณ
SAS
โˆ ๐‘ฑ๐‘น๐‘ฒ โ‰… โˆ ๐‘ฑ๐‘น๐‘ณ
Corresponding angles of congruent triangles are congruent.
๐’Žโˆ ๐‘ฑ๐‘น๐‘ฒ + ๐’Žโˆ ๐‘ฑ๐‘น๐‘ณ = ๐Ÿ๐Ÿ–๐ŸŽ°
Linear pairs form supplementary angles.
๐Ÿ(๐’Žโˆ ๐‘ฑ๐‘น๐‘ฒ) = ๐Ÿ๐Ÿ–๐ŸŽ°
Substitution property of equality
๐’Žโˆ ๐‘ฑ๐‘น๐‘ฒ = ๐Ÿ—๐ŸŽ°
Division property of equality
ฬ…ฬ…ฬ…ฬ…
โˆด ฬ…ฬ…ฬ…ฬ…
๐‘ฑ๐‘น โŠฅ ๐‘ฒ๐‘ณ
Definition of perpendicular lines
Lesson 23:
Base Angles of Isosceles Triangles
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M1-TE-1.3.0-07.2015
202
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 23
NYS COMMON CORE MATHEMATICS CURRICULUM
M1
GEOMETRY
2.
๐‘จ๐‘ฉ = ๐‘จ๐‘ช, ๐‘ฟ๐‘ฉ = ๐‘ฟ๐‘ช
ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ฟ bisects โˆ ๐‘ฉ๐‘จ๐‘ช
Given:
Prove:
๐‘จ๐‘ฉ = ๐‘จ๐‘ช
Given
๐’Žโˆ ๐‘จ๐‘ฉ๐‘ช = ๐’Žโˆ ๐‘จ๐‘ช๐‘ฉ
Base angles of an isosceles triangle are equal in
measure.
๐‘ฟ๐‘ฉ = ๐‘ฟ๐‘ช
Given
๐’Žโˆ ๐‘ฟ๐‘ฉ๐‘ช = ๐’Žโˆ ๐‘ฟ๐‘ช๐‘ฉ
Base angles of an isosceles triangle are equal in
measure.
๐’Žโˆ ๐‘จ๐‘ฉ๐‘ช = ๐’Žโˆ ๐‘จ๐‘ฉ๐‘ฟ + ๐’Žโˆ ๐‘ฟ๐‘ฉ๐‘ช,
๐’Žโˆ ๐‘จ๐‘ช๐‘ฉ = ๐’Žโˆ ๐‘จ๐‘ช๐‘ฟ + ๐’Žโˆ ๐‘ฟ๐‘ช๐‘ฉ
Angle addition postulate
๐’Žโˆ ๐‘จ๐‘ฉ๐‘ฟ = ๐’Žโˆ ๐‘จ๐‘ฉ๐‘ช โˆ’ ๐’Žโˆ ๐‘ฟ๐‘ฉ๐‘ช,
3.
๐’Žโˆ ๐‘จ๐‘ช๐‘ฟ = ๐’Žโˆ ๐‘จ๐‘ช๐‘ฉ โˆ’ ๐’Žโˆ ๐‘ฟ๐‘ช๐‘ฉ
Subtraction property of equality
๐’Žโˆ ๐‘จ๐‘ฉ๐‘ฟ = ๐’Žโˆ ๐‘จ๐‘ช๐‘ฟ
Substitution property of equality
โˆด โ–ณ ๐‘จ๐‘ฉ๐‘ฟ โ‰… โ–ณ ๐‘จ๐‘ช๐‘ฟ
SAS
โˆ ๐‘ฉ๐‘จ๐‘ฟ โ‰… โˆ ๐‘ช๐‘จ๐‘ฟ
Corresponding parts of congruent triangles are congruent.
โˆด ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ฟ bisects โˆ ๐‘ฉ๐‘จ๐‘ช.
Definition of an angle bisector
Given:
๐‘ฑ๐‘ฟ = ๐‘ฑ๐’€, ๐‘ฒ๐‘ฟ = ๐‘ณ๐’€
Prove:
โ–ณ ๐‘ฑ๐‘ฒ๐‘ณ is isosceles
๐‘ฑ๐‘ฟ = ๐‘ฑ๐’€
Given
๐‘ฒ๐‘ฟ = ๐‘ณ๐’€
Given
๐’Žโˆ ๐Ÿ = ๐’Žโˆ ๐Ÿ
Base angles of an isosceles triangle are
equal in measure.
๐’Žโˆ ๐Ÿ + ๐’Žโˆ ๐Ÿ‘ = ๐Ÿ๐Ÿ–๐ŸŽ°
Linear pairs form supplementary angles.
๐’Žโˆ ๐Ÿ + ๐’Žโˆ ๐Ÿ’ = ๐Ÿ๐Ÿ–๐ŸŽ°
Linear pairs form supplementary angles.
๐’Žโˆ ๐Ÿ‘ = ๐Ÿ๐Ÿ–๐ŸŽ° โˆ’ ๐’Žโˆ ๐Ÿ
Subtraction property of equality
๐’Žโˆ ๐Ÿ’ = ๐Ÿ๐Ÿ–๐ŸŽ° โˆ’ ๐’Žโˆ ๐Ÿ‘
Subtraction property of equality
๐’Žโˆ ๐Ÿ‘ = ๐’Žโˆ ๐Ÿ’
Substitution property of equality
โˆด โ–ณ ๐‘ฑ๐‘ฒ๐‘ฟ โ‰…โ–ณ ๐‘ฑ๐’€๐‘ณ
SAS
ฬ…ฬ…ฬ…ฬ…
๐‘ฑ๐‘ฒ โ‰… ฬ…ฬ…ฬ…
๐‘ฑ๐‘ณ
Corresponding parts of congruent triangles are congruent.
โˆด โ–ณ ๐‘ฑ๐‘ฒ๐‘ณ is isosceles.
Definition of an isosceles triangle
Lesson 23:
Base Angles of Isosceles Triangles
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M1-TE-1.3.0-07.2015
203
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 23
NYS COMMON CORE MATHEMATICS CURRICULUM
M1
GEOMETRY
4.
Given:
โ–ณ ๐‘จ๐‘ฉ๐‘ช, with ๐’Žโˆ ๐‘ช๐‘ฉ๐‘จ = ๐’Žโˆ ๐‘ฉ๐‘ช๐‘จ
Prove:
๐‘ฉ๐‘จ = ๐‘ช๐‘จ
(Converse of base angles of isosceles triangle)
Hint: Use a transformation.
PROOF:
We can prove that ๐‘จ๐‘ฉ = ๐‘จ๐‘ช by using rigid motions. Construct the perpendicular bisector
ฬ…ฬ…ฬ…ฬ…. Note that we do not know whether the point ๐‘จ is on ๐’. If we did, we would
๐’ to ๐‘ฉ๐‘ช
know immediately that ๐‘จ๐‘ฉ = ๐‘จ๐‘ช, since all points on the perpendicular bisector are
equidistant from ๐‘ฉ and ๐‘ช. We need to show that ๐‘จ is on ๐’, or equivalently, that the
reflection across ๐’ takes ๐‘จ to ๐‘จ.
Let ๐’“๐’ be the transformation that reflects the โ–ณ ๐‘จ๐‘ฉ๐‘ช across ๐’. Since ๐’ is the perpendicular bisector, ๐’“๐’ (๐‘ฉ) = ๐‘ช and
๐’“๐’ (๐‘ช) = ๐‘ฉ. We do not know where the transformation takes ๐‘จ; for now, let us say that ๐’“๐’ (๐‘จ) = ๐‘จโ€ฒ .
Since โˆ ๐‘ช๐‘ฉ๐‘จ โ‰… โˆ ๐‘ฉ๐‘ช๐‘จ and rigid motions preserve angle measures, after applying ๐’“๐’ to โˆ ๐‘ฉ๐‘ช๐‘จ, we get that
โˆ ๐‘ช๐‘ฉ๐‘จ โ‰… โˆ ๐‘ช๐‘ฉ๐‘จโ€ฒ . โˆ ๐‘ช๐‘ฉ๐‘จ and โˆ ๐‘ช๐‘ฉ๐‘จโ€ฒ share a ray ๐‘ฉ๐‘ช, are of equal measure, and ๐‘จ and ๐‘จโ€ฒ are both in the same halfโƒ—โƒ—โƒ—โƒ—โƒ—โƒ— and โƒ—โƒ—โƒ—โƒ—โƒ—โƒ—โƒ—
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— .
plane with respect to line ๐‘ฉ๐‘ช. Hence, ๐‘ฉ๐‘จ
๐‘ฉ๐‘จโ€ฒ are the same ray. In particular, ๐‘จโ€ฒ is a point somewhere on ๐‘ฉ๐‘จ
Likewise, applying ๐’“๐’ to โˆ ๐‘ช๐‘ฉ๐‘จ gives โˆ ๐‘ฉ๐‘ช๐‘จ โ‰… โˆ ๐‘ฉ๐‘ช๐‘จโ€ฒ , and for the same reasons in the previous paragraph, ๐‘จโ€ฒ must
โƒ—โƒ—โƒ—โƒ—โƒ— . Therefore, ๐‘จโ€ฒ is common to both ๐‘ฉ๐‘จ
โƒ—โƒ—โƒ—โƒ—โƒ— .
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— and ๐‘ช๐‘จ
also be a point somewhere on ๐‘ช๐‘จ
The only point common to both โƒ—โƒ—โƒ—โƒ—โƒ—โƒ—
๐‘ฉ๐‘จ and โƒ—โƒ—โƒ—โƒ—โƒ—
๐‘ช๐‘จ is point ๐‘จ; thus, ๐‘จ and ๐‘จโ€ฒ must be the same point, i.e., ๐‘จ = ๐‘จโ€ฒ .
Hence, the reflection takes ๐‘จ to ๐‘จ, which means ๐‘จ is on the line ๐’ and ๐’“๐’ (๐‘ฉ๐‘จ) = ๐‘ช๐‘จโ€ฒ = ๐‘ช๐‘จ, or ๐‘ฉ๐‘จ = ๐‘ช๐‘จ.
5.
Given:
ฬ…ฬ…ฬ…ฬ… โˆฅ ๐‘ฟ๐’€
ฬ…ฬ…ฬ…ฬ…ฬ…is the angle bisector of โˆ ๐‘ฉ๐’€๐‘จ, and ๐‘ฉ๐‘ช
ฬ…ฬ…ฬ…ฬ…
โ–ณ ๐‘จ๐‘ฉ๐‘ช, with ๐‘ฟ๐’€
Prove:
๐’€๐‘ฉ = ๐’€๐‘ช
ฬ…ฬ…ฬ…ฬ…
ฬ…ฬ…ฬ…ฬ…
๐‘ฉ๐‘ช โˆฅ ๐‘ฟ๐’€
Given
๐’Žโˆ ๐‘ฟ๐’€๐‘ฉ = ๐’Žโˆ ๐‘ช๐‘ฉ๐’€
When two parallel lines are cut by
a transversal, the alternate
interior angles are equal.
๐’Žโˆ ๐‘ฟ๐’€๐‘จ = ๐’Žโˆ ๐‘ฉ๐‘ช๐’€
When two parallel lines are cut by
a transversal, the corresponding
angles are equal.
๐’Žโˆ ๐‘ฟ๐’€๐‘จ = ๐’Žโˆ ๐‘ฟ๐’€๐‘ฉ
Definition of an angle bisector
๐’Žโˆ ๐‘ช๐‘ฉ๐’€ = ๐’Žโˆ ๐‘ฉ๐‘ช๐’€
Substitution property of equality
๐’€๐‘ฉ = ๐’€๐‘ช
When the base angles of a triangle are equal in measure, the triangle is
isosceles.
Closing (1 minute)
๏‚ง
Would you prefer to prove that the base angles of an isosceles triangle are equal in measure using
transformations or by using SAS? Why?
Exit Ticket (3 minutes)
Lesson 23:
Base Angles of Isosceles Triangles
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Lesson 23
NYS COMMON CORE MATHEMATICS CURRICULUM
M1
GEOMETRY
Name
Date
Lesson 23: Base Angles of Isosceles Triangles
Exit Ticket
For each of the following, if the given congruence exists, name the isosceles triangle and the pair of congruent angles for
the triangle based on the image above.
1.
ฬ…ฬ…ฬ…ฬ…
ฬ…ฬ…ฬ…ฬ…
๐ด๐ธ โ‰… ๐ฟ๐ธ
2.
ฬ…ฬ…ฬ…ฬ… โ‰… ฬ…ฬ…ฬ…ฬ…
๐ฟ๐ธ
๐ฟ๐บ
3.
ฬ…ฬ…ฬ…ฬ… โ‰… ๐ฟ๐‘
ฬ…ฬ…ฬ…ฬ…
๐ด๐‘
4.
ฬ…ฬ…ฬ…ฬ… โ‰… ฬ…ฬ…ฬ…ฬ…
๐ธ๐‘
๐บ๐‘
5.
ฬ…ฬ…ฬ…ฬ…
๐‘๐บ โ‰… ฬ…ฬ…ฬ…ฬ…
๐ฟ๐บ
6.
ฬ…ฬ…ฬ…ฬ…
๐ด๐ธ โ‰… ฬ…ฬ…ฬ…ฬ…
๐‘๐ธ
Lesson 23:
Base Angles of Isosceles Triangles
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M1-TE-1.3.0-07.2015
205
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Lesson 23
NYS COMMON CORE MATHEMATICS CURRICULUM
M1
GEOMETRY
Exit Ticket Sample Solutions
For each of the following, if the given congruence exists, name the isosceles triangle and the pair of congruent angles for
the triangle based on the image above.
1.
ฬ…ฬ…ฬ…ฬ… โ‰… ๐‘ณ๐‘ฌ
ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ฌ
2.
โ–ณ ๐‘จ๐‘ฌ๐‘ณ, โˆ ๐‘ฌ๐‘จ๐‘ณ โ‰… โˆ ๐‘ฌ๐‘ณ๐‘จ
3.
โ–ณ ๐‘ณ๐‘ฌ๐‘ฎ, โˆ ๐‘ณ๐‘ฌ๐‘ฎ โ‰… โˆ ๐‘ณ๐‘ฎ๐‘ฌ
ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ต โ‰… ฬ…ฬ…ฬ…ฬ…
๐‘ณ๐‘ต
4.
โ–ณ ๐‘ต๐‘ณ๐‘จ, โˆ ๐‘ต๐‘ณ๐‘จ โ‰… โˆ ๐‘ต๐‘จ๐‘ณ
5.
ฬ…ฬ…ฬ…ฬ…
ฬ…ฬ…ฬ…ฬ… โ‰… ๐‘ณ๐‘ฎ
๐‘ณ๐‘ฌ
ฬ…ฬ…ฬ…ฬ…
๐‘ฌ๐‘ต โ‰… ฬ…ฬ…ฬ…ฬ…ฬ…
๐‘ฎ๐‘ต
โ–ณ ๐‘ต๐‘ฎ๐‘ฌ, โˆ ๐‘ต๐‘ฎ๐‘ฌ โ‰… โˆ ๐‘ต๐‘ฌ๐‘ฎ
ฬ…ฬ…ฬ…ฬ…ฬ… โ‰… ๐‘ณ๐‘ฎ
ฬ…ฬ…ฬ…ฬ…
๐‘ต๐‘ฎ
6.
โ–ณ ๐‘ฎ๐‘ณ๐‘ต, โˆ ๐‘ฎ๐‘ต๐‘ณ โ‰… โˆ ๐‘ฎ๐‘ณ๐‘ต
ฬ…ฬ…ฬ…ฬ… โ‰… ๐‘ต๐‘ฌ
ฬ…ฬ…ฬ…ฬ…
๐‘จ๐‘ฌ
โ–ณ ๐‘จ๐‘ฌ๐‘ต, โˆ ๐‘ฌ๐‘ต๐‘จ โ‰… โˆ ๐‘ฌ๐‘จ๐‘ต
Problem Set Sample Solutions
1.
Given:
๐‘จ๐‘ฉ = ๐‘ฉ๐‘ช, ๐‘จ๐‘ซ = ๐‘ซ๐‘ช
Prove:
โ–ณ ๐‘จ๐‘ซ๐‘ฉ and โ–ณ ๐‘ช๐‘ซ๐‘ฉ are right triangles
๐‘จ๐‘ฉ = ๐‘ฉ๐‘ช
Given
โ–ณ ๐‘จ๐‘ฉ๐‘ช is isosceles.
Definition of isosceles triangle
๐’Žโˆ ๐‘จ = ๐’Žโˆ ๐‘ช
Base angles of an isosceles triangle
are equal in measure.
๐‘จ๐‘ซ = ๐‘ซ๐‘ช
Given
โ–ณ ๐‘จ๐‘ซ๐‘ฉ โ‰… โ–ณ ๐‘ช๐‘ซ๐‘ฉ
SAS
๐’Žโˆ ๐‘จ๐‘ซ๐‘ฉ = ๐’Žโˆ ๐‘ช๐‘ซ๐‘ฉ
Corresponding angles of congruent
triangles are equal in measure.
๐’Žโˆ ๐‘จ๐‘ซ๐‘ฉ + ๐’Žโˆ ๐‘ช๐‘ซ๐‘ฉ = ๐Ÿ๐Ÿ–๐ŸŽ°
Linear pairs form supplementary angles.
๐Ÿ(๐’Žโˆ ๐‘จ๐‘ซ๐‘ฉ) = ๐Ÿ๐Ÿ–๐ŸŽ°
Substitution property of equality
๐’Žโˆ ๐‘จ๐‘ซ๐‘ฉ = ๐Ÿ—๐ŸŽ°
Division property of equality
๐’Žโˆ ๐‘ช๐‘ซ๐‘ฉ = ๐Ÿ—๐ŸŽ°
Transitive property
โ–ณ ๐‘จ๐‘ซ๐‘ฉ and โ–ณ ๐‘ช๐‘ซ๐‘ฉ are right triangles.
Definition of a right triangle
Lesson 23:
Base Angles of Isosceles Triangles
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M1-TE-1.3.0-07.2015
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This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 23
M1
GEOMETRY
2.
3.
Given:
ฬ…ฬ…ฬ…ฬ…
ฬ…ฬ…ฬ…ฬ… โ€–๐‘ช๐‘ฌ
๐‘จ๐‘ช = ๐‘จ๐‘ฌ and ๐‘ฉ๐‘ญ
Prove:
๐‘จ๐‘ฉ = ๐‘จ๐‘ญ
๐‘จ๐‘ช = ๐‘จ๐‘ฌ
Given
โ–ณ ๐‘จ๐‘ช๐‘ฌ is isosceles.
Definition of isosceles triangle
๐’Žโˆ ๐‘จ๐‘ช๐‘ฌ = ๐’Žโˆ ๐‘จ๐‘ฌ๐‘ช
Base angles of an isosceles triangle are
equal in measure.
ฬ…ฬ…ฬ…ฬ…
๐‘ฉ๐‘ญ โ€– ฬ…ฬ…ฬ…ฬ…
๐‘ช๐‘ฌ
Given
๐’Žโˆ ๐‘จ๐‘ช๐‘ฌ = ๐’Žโˆ ๐‘จ๐‘ฉ๐‘ญ
If parallel lines are cut by a transversal,
then corresponding angles are equal in
measure.
๐’Žโˆ ๐‘จ๐‘ญ๐‘ฉ = ๐’Žโˆ ๐‘จ๐‘ฌ๐‘ช
If parallel lines are cut by a transversal,
then corresponding angles are equal in measure.
๐’Žโˆ ๐‘จ๐‘ญ๐‘ฉ = ๐’Žโˆ ๐‘จ๐‘ฉ๐‘ญ
Transitive property
โ–ณ ๐‘จ๐‘ฉ๐‘ญ is isosceles.
If the base angles of a triangle are equal in measure, the triangle is isosceles.
๐‘จ๐‘ฉ = ๐‘จ๐‘ญ
Definition of an isosceles triangle
ฬ…ฬ…ฬ…ฬ… โ‰… ๐‘จ๐‘ฉ
ฬ…ฬ…ฬ…ฬ…. In your own words, describe how transformations and the
In the diagram, โ–ณ ๐‘จ๐‘ฉ๐‘ช is isosceles with ๐‘จ๐‘ช
properties of rigid motions can be used to show that โˆ ๐‘ช โ‰… โˆ ๐‘ฉ.
Answers will vary. A possible response would summarize the key
elements of the proof from the lesson, such as the response below.
It can be shown that โˆ ๐‘ช โ‰… โˆ ๐‘ฉ using a reflection across the bisector of โˆ ๐‘จ.
The two angles formed by the bisector of โˆ ๐‘จ would be mapped onto one
another since they share a ray, and rigid motions preserve angle measure.
Since segment length is also preserved, ๐‘ฉ is mapped onto ๐‘ช and vice
versa.
Finally, since rays are taken to rays and rigid motions preserve angle
โƒ—โƒ—โƒ—โƒ—โƒ— is mapped onto ๐‘ฉ๐‘จ
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— is mapped onto ๐‘ฉ๐‘ช
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— , and โˆ ๐‘ช is
โƒ—โƒ—โƒ—โƒ—โƒ—โƒ— , ๐‘ช๐‘ฉ
measure, ๐‘ช๐‘จ
mapped onto โˆ ๐‘ฉ. From this, we see that โˆ ๐‘ช โ‰… โˆ ๐‘ฉ.
Lesson 23:
Base Angles of Isosceles Triangles
This work is derived from Eureka Math โ„ข and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from GEO-M1-TE-1.3.0-07.2015
207
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.