Chapter 6.4, Problem 2AE

To determine

To show: line segment $\overline{AS}\text{ and }\overline{BS}$ reflected paths of light and to prove $\angle QSA\cong \angle PSB$

Explanation of Solution

Given: point $B$ is the reflecting to the line $l$ and $S$ is a solution by reflection.

  $\angle BPS={90}^{\circ }$

Since point $B$ is the reflecting to the line $l$ and $S$ is a solution by reflection we can tell that line segment $\overline{AS}\text{ and }\overline{BS}$ reflected paths of light.

Construction: Join line $QA$

  

Proof: In $\Delta BPS\text{ and }\Delta \text{AQS}$

If two angles are complementary to two other congruent angles, then they are congruent.

Clearly, we can see that in two triangle angles $\angle QSA$ , $\angle BPS$ are complementary angles.

Therefore, $\angle QSA\cong \angle PSB$