   Chapter 5.3, Problem 15E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

In Exercises 13 to 16, provide the missing reasons.Given: Δ A B C ;   M   and  N are midpoints of A B ¯ and A C ¯ , respectivelyProve: Δ A M N ∼ Δ A B C PROOF Statements Reasons 1. Δ A B C ;   M   and  N are the midpoints of A B ¯ and A C ¯ , respectively. 1. ? 2. A M = 1 2 ( A B ) and A N = 1 2 ( A C ) 2. ? 3. M N = 1 2 ( B C ) 3. ? 4. A M A B = 1 2 , A N A C = 1 2 , and  M N B C = 1 2 4. ? 5. A M A B = A N A C = M N B C 5.? 6. Δ A M N ∼ Δ A B C 6. ?

To determine

To provide:

The missing reasons.

Explanation

Given:

Given: ΔABC; M and N are midpoints of AB¯ and AC¯, respectively.

Prove: ΔAMNΔABC

 PROOF Statements Reasons 1. ΔABC; M and N are midpoints of AB¯ and AC¯, respectively. 1. ? 2. AM=12(AB) andAN=12(AC) 2. ? 3. MN=12(BC) 3. ? 4. AMAB=12,ANAC=12,and MNBC=12 4. ? 5. AMAB=ANAC=MNBC 5.? 6. ΔAMN∼ΔABC 6. ?

Definition:

SSS:

If the three sides of one triangle are proportional (in length) to the three corresponding sides of a second triangle, then the triangle are similar.

Description:

Given that ΔABC;Mand N are midpoints of AB¯ and AC¯, respectively.

To Prove: ΔAMNΔABC

From the given figure, it is observed that the line segment MN joining the two sides of AB and AC. In the triangle ABC, M is the midpoint of AB and N is the midpoint of AC.

By the definition of midpoint, AM=12AB and AN=12AC.

It is known that the line segment MN joins the midpoints of two sides a triangle, its length MN is half the length of BC

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