Question

In Exercise 18-20, refer to the diagrams on page 578. Given the reflection $R_m:\overrightarrow{PQ}\to \overrightarrow{P\text{'}Q\text{'}}$, write the key steps of a proof that $PQ=P\text{'}Q\text{'}$for each case.

Case 2.

Short answer

The required proof is as shown in the below steps.

Step-by-step solution

Step 1. Given Information.

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

Step 2. Formula Used.

Definition of reflection.

Step 3. 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.

Step 4. Conclusion.

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