# Eureka Math Geometry Module 1 Lesson 27 Answer Key

## Engage NY Eureka Math Geometry Module 1 Lesson 27 Answer Key

### Eureka Math Geometry Module 1 Lesson 27 Exercise Answer Key

Exercises

Exercise 1.
Given: AB=AC, RB=RC
Prove: SB=SC

AB=AC, RB=RC Given
AR=AR Reflexive property
△ARC≅ △ARB SSS
m∠ARC=m∠ARB Corresponding angles of congruent triangles are equal in measure.
m∠ARC+m∠SRC=180°,
m∠ARB+m∠SRB=180° Linear pairs form supplementary angles.
m∠SRC=m∠SRB Angles supplementary to either the same angle or to congruent angles are equal in measure.
SR=RS Reflexive property
△SRB≅ △SRC SAS
SB=SC Corresponding sides of congruent angles are equal in length.

Exercise 2.
Given: Square ABCS ≅ Square EFGS,
$$\overleftrightarrow{\boldsymbol{R A B}}$$, $$\overleftrightarrow{\boldsymbol{R E F}}$$
Prove: △ASR≅ △ESR

Square ABCS ≅ Square EFGS Given
AS=ES Corresponding sides of congruent squares are equal in length.
SR=RS Reflexive property
∠BAS and ∠FES are right angles. Definition of square
∠BAS and ∠SAR form a linear pair. Definition of linear pair
∠FES and ∠SER form a linear pair. Definition of linear pair
∠SAR and ∠SER are right angles. Two angles that are supplementary and congruent each measure 90° and are, therefore, right angles.
△ASR and △ESR are right triangles. Definition of right triangle
△ASR≅ △ESR HL

Exercise 3.
Given: JK=JL, JX=JY
Prove: KX=LY

JX=JY Given
m∠JXY=m∠JYX Base angles of an isosceles triangle are equal in measure.
m∠JXK+m∠JXY=180°,
m∠JYL+m∠JYX=180° Linear pairs form supplementary angles.
m∠JXK+m∠JXY=m∠JYL+m∠JYX Substitution property of equality
m∠JXK+m∠JXY=m∠JYL+m∠JXY Substitution property of equality
m∠JXK=m∠JYL Angles supplementary to either the same angle or congruent angles are equal in measure.
JK=JL Given
m∠K=m∠L Base angles of an isosceles triangle are equal in measure.
△JXK≅ △JYL AAS
KX=LY Corresponding sides of congruent triangles are equal in length.

Exercise 4.
Given: $$\overline{A D}$$ ⊥$$\overline{D R}$$, $$\overline{A B}$$ ⊥$$\overline{B R}$$,
$$\overline{A D}$$ ≅ $$\overline{A B}$$
Prove: ∠DCR≅∠BCR

$$\overline{A D}$$ ⊥$$\overline{D R}$$, $$\overline{A B}$$ ⊥$$\overline{B R}$$ Given
△ADR and △ABR are right triangles. Definition of right triangle
$$\overline{A D}$$ ≅ $$\overline{A B}$$ Given
$$\overline{A R}$$ ≅ $$\overline{A R}$$ Reflexive property
∠ARD≅ARB Corresponding angles of congruent triangles are congruent.
m∠ARD+m∠DRC=180°,
m∠ARB+m∠BRC=180° Linear pairs form supplementary angles.
m∠ARD+m∠DRC=m∠ARB+m∠BRC Transitive property
m∠DRC=m∠BRC Angles supplementary to either the same angle or congruent angles are equal in measure.
$$\overline{D R}$$ ≅ $$\overline{B R}$$ Corresponding sides of congruent triangles are congruent.
$$\overline{R C}$$ ≅$$\overline{R C}$$ Reflexive property
△DRC≅ △BRC SAS
∠DRC≅∠BRC Corresponding angles of congruent triangles are congruent.

Exercise 5.
Given: AR=AS, BR=CS,
$$\overline{R X}$$ ⊥$$\overline{A B}$$, $$\overline{S Y}$$ ⊥$$\overline{A C}$$
Prove: BX=CY

AR=AS Given
m∠ARS=m∠ASR Base angles of an isosceles triangle are equal in measure.
m∠ARS+m∠ARB=180°,
m∠ASR+m∠ASC=180° Linear pairs form supplementary angles.
m∠ARS+m∠ARB=m∠ASR+m∠ASC Transitive property
m∠ARB=m∠ASC Subtraction
BR=CS Given
△ARB≅ △ASC SAS
∠ABR≅∠ACS Corresponding angles of congruent triangles are congruent.
$$\overline{R X}$$ ⊥$$\overline{A B}$$, $$\overline{S Y}$$ ⊥$$\overline{A C}$$ Given
m∠RXB=90°=m∠SYC Definition of perpendicular line segments
△BRX≅ △SYC AAS
BX=CY Corresponding sides of congruent triangles are equal in length.

Exercise 6.
Given: AX=BX, m∠AMB=m∠AYZ=90°
Prove: NY=NM

AX=BX Given
m∠AMB=m∠AYZ=90° Given
m∠BXM=m∠AXY Vertical angles are equal in measure.
△BXM≅△AXY AAS
XM=XY Corresponding sides of congruent triangles are equal in length.
BY=AM Substitution property of equality
m∠BYN=90° Vertical angles are equal in measure.
m∠AMB+m∠AMN=180° Linear pairs form supplementary angles.
m∠AMN=90° Subtraction property of equality
m∠MNY=m∠MNY Reflexive property
△BYN≅ △AMN AAS
NY=NM Corresponding sides of congruent triangles are equal in length.

### Eureka Math Geometry Module 1 Lesson 27 Problem Set Answer Key

Use your knowledge of triangle congruence criteria to write a proof for the following:
In the figure $$\overline{B E}$$ ≅ $$\overline{C E}$$, $$\overline{D C}$$ ⊥$$\overline{A B}$$, and $$\overline{B E}$$ ⊥$$\overline{A C}$$ ; prove $$\overline{A E}$$ ≅ $$\overline{R E}$$.

m∠ERC=m∠BRD Vertical angles are equal in measure.
$$\overline{D C}$$ ⊥$$\overline{A B}$$, $$\overline{B E}$$ ⊥$$\overline{A C}$$ Given
m∠BDR=90°, m∠REC=90° Definition of perpendicular lines
m∠ABE=m∠RCE Sum of the angle measures in a triangle is 180°.
m∠BAE=m∠BRD Sum of the angle measures in a triangle is 180°.
m∠BAE=m∠ERC Substitution property of equality
$$\overline{B E}$$ ≅ $$\overline{C E}$$ Given
△BAE≅ △CRE AAS
$$\overline{A E}$$ ≅$$\overline{R E}$$ Corresponding sides of congruent triangles are congruent.

### Eureka Math Geometry Module 1 Lesson 27 Exit Ticket Answer Key

Given: M is the midpoint of $$\overline{G R}$$, ∠G ≅∠R
Prove: △GHM≅ △RPM

M is the midpoint of $$\overline{G R}$$. Given