6.PracticeCRM 2.1 - Lesson 14Consider the figure below.MABCGiven: mLA = mzD, AP = DQ, AB = BC = CDProve: BM = CMa. Complete the following two-column proof.StatementsReasons1. mLA = mzD1. Given2. AP = DQ, AB = BC = CD2. Given3.3. Segment Addition Postulate4. AB + BC = DB4.5.5. Transitive Property6. A CAP =A BDQ6.7.7. СРСТС8. BM = CM8.b.Prove BM = CM by applying properties oftransformations. Justify your steps.

Given: m∠A=m∠D, AP¯≅DQ¯, AB¯≅BC¯≅CD¯

Consider the figure below. P M A B Given: mLA = mLD, AP = DQ, AB = BC = CD Prove: BM = CM a. Complete the following two-column proof. Statements Reasons 1. mLA = m2D 1. Given 2. AP = DQ, AB = BC = CD 2. Given 3. 3. Segment Addition Postulate 4. AB + BC = DB 4. 5. 5. Transitive Property 6. △ CAP 열스 BDQ 6. 7. |7. СРСТС 8. BM = CM 8. b. Prove BM = CM by applying properties of transformations. Justify your steps.

Consider the figure below. P M A B Given: mLA = mLD, AP = DQ, AB = BC = CD Prove: BM = CM a. Complete the following two-column proof. Statements Reasons 1. mLA = m2D 1. Given 2. AP = DQ, AB = BC = CD 2. Given 3. 3. Segment Addition Postulate 4. AB + BC = DB 4. 5. 5. Transitive Property 6. △ CAP 열스 BDQ 6. 7. |7. СРСТС 8. BM = CM 8. b. Prove BM = CM by applying properties of transformations. Justify your steps.

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterP: Preliminary Concepts

SectionP.CT: Test

Problem 1CT

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Question

6.
Practice
CRM 2.1 - Lesson 14
Consider the figure below.
M
A
B
C
Given: mLA = mzD, AP = DQ, AB = BC = CD
Prove: BM = CM
a. Complete the following two-column proof.
Statements
Reasons
1. mLA = mzD
1. Given
2. AP = DQ, AB = BC = CD
2. Given
3.
3. Segment Addition Postulate
4. AB + BC = DB
4.
5.
5. Transitive Property
6. A CAP =A BDQ
6.
7.
7. СРСТС
8. BM = CM
8.
b.
Prove BM = CM by applying properties of
transformations. Justify your steps.

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