Chapter 14.2, Problem 18WE

To determine

To prove: $PQ=P\text{'}Q\text{'}$

Explanation of Solution

Given information: given the reflection $R_m=\overline{PQ}\to \overline{P\text{'}Q\text{'}}$ and the diagram,

Formula used: definition of reflection.

Proof:

In the given reflection through m $P=P\text{'}$

And $Q$ has a reflection $Q\text{'}$ so from the definition of reflection we know that $\overline{QQ\text{'}}\perp m$ and, $Q$ and $Q\text{'}$ are equidistance from line m, and let O be the intersection of $\overline{QQ\text{'}}$ and $m$ .

Here

$PO$ is common, $\angle POQ=\angle POQ\text{'}$ and $OQ=OQ\text{'}$

Hence $\Delta POQ\widetilde{=}\Delta POQ\text{'}$

Now as the triangle are congruent,it can be concluded that $PQ=P\text{'}Q\text{'}$ .

Therefore, $PQ=P\text{'}Q\text{'}$ is proved.