Chapter 6.1, Problem 17PSC

To determine

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

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 .

  $Xison\overrightarrow{AB}.$

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 .

  $Xison\overrightarrow{AB}.$

P and Q are above m .

B is at equal distance from P and Q.

A is at equal distance from P and Q.

  

Points distant from same point form congruent segments.

  $\begin{array}{l}\overline{PB} \cong \overline{BQ}\\ \overline{PA} \cong \overline{QA}\end{array}$

By reflexive property, we get

  $\overline{BA} \cong \overline{BA}$

By SSS congruence rule, we get

  $\Delta PAB \cong \Delta QAB$

As corresponding parts of congruent triangles are congruent, we get

  $\angle PAX\cong \angle QAX$

By reflexive property, we get

  $\overline{AX} \cong \overline{AX}$

By SAS congruence rule, we get

  $\Delta PAX \cong \Delta QAX$

As corresponding parts of congruent triangles are congruent, we get

  $\overline{PX}\cong \overline{QX}$

According to definition of equal distance,

X is equal distance from P and Q.