Chapter 3.1, Problem 41E

### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

Chapter
Section

### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# Given: âˆ  A B C ; R S Â¯ is the perpendicular bisector of A B Â¯ ; R T Â¯ is theperpendicular bisector of B C Â¯ . Prove: A R Â¯ â‰… R C Â¯

To determine

To prove:

The segment AR¯ is congruent to the segment RC¯.

Explanation

Given:

The given figure is,

Figure (1)

From figure (1).

RSÂ¯ is the perpendicular bisector of ABÂ¯

And,

RTÂ¯ is the perpendicular bisector of BCÂ¯.

Concept:

The perpendicular bisector bisect the segment in two congruent part.

The perpendicular bisector makes an angle of 90Â° with the bisected segment.

Approach:

Consider the triangle,

Î”RTB and Î”RTC,

Since, RTÂ¯ is the perpendicular bisector. Then,

BTÂ¯â‰…TCÂ¯âˆ RTBâ‰…âˆ RTC

Since, the side RTÂ¯ is common in both the triangles Î”RTB and Î”RTC. Then,

RTÂ¯â‰…RTÂ¯

Thus, from SAS property if two sides and one angle are congruent in two triangles, then

Î”RTBâ‰…Î”RTC

Since, the congruent triangles have congruent corresponding segments. Then,

RBÂ¯â‰…RCÂ¯ â€¦â€¦(1)

Consider the triangle,

Î”RSA and Î”RSB,

Since, RSÂ¯ is the perpendicular bisector

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