Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
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Chapter 6.1, Problem 17PSC
To determine

To show that point X is equidistant from points P and Q.

Expert Solution & Answer
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Answer to Problem 17PSC

Point X is equidistant from points P and Q.

Explanation of Solution

Given information:

A, X and B lie in plane m .

  X is on AB.

P and Q are above m .

B is at equal distance from P and Q.

A is at equal distance from P and Q.

Formula used:

The below properties are used:

Points distant from same point form congruent segments.

Two triangles are congruent by SSS congruence rule.

Two triangles are congruent by SAS congruence rule.

Proof:

It is given that,

A, X and B lie in plane m .

  X is on AB.

P and Q are above m .

B is at equal distance from P and Q.

A is at equal distance from P and Q.

  Geometry For Enjoyment And Challenge, Chapter 6.1, Problem 17PSC

Points distant from same point form congruent segments.

  PB¯   BQ¯PA¯   QA¯

By reflexive property, we get

  BA¯   BA¯

By SSS congruence rule, we get

  ΔPAB    ΔQAB  

As corresponding parts of congruent triangles are congruent, we get

  PAX   QAX

By reflexive property, we get

  AX¯   AX¯

By SAS congruence rule, we get

  ΔPAX    ΔQAX  

As corresponding parts of congruent triangles are congruent, we get

  PX¯   QX¯

According to definition of equal distance,

X is equal distance from P and Q.

Chapter 6 Solutions

Geometry For Enjoyment And Challenge

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