Chapter 14.2, Problem 24WE

a.

To determine

To prove that $\angle 1\cong \angle 2,\angle 1\cong \angle 3\text{and}\angle 3\cong \angle 2$ .

Explanation of Solution

Given information:

$H\text{'}$ Is the image of H by reflection in the wall and $\overline{BH\text{'}}$ intersects the wall at P .

Proof:

$\angle 1\cong \angle 3\left[\text{vertically opposite angles}\right]$

In triangles $PXH\text{ and }PXH\text{'}$

$\begin{array}{l}PX=PX\left[\text{common side}\right]\\ m\angle X=m\angle X\left[\text{each equals to 90}°\right]\\ HX=HX\text{'}\left[\text{perpendicular bisector}\right]\end{array}$

By side angle side congruency

$\begin{array}{l}\Delta PXH\cong \Delta PXH\text{'}\\ \angle 2\cong \angle 3\end{array}$

Since $\angle 1\cong \angle 3\text{ and }\angle 2\cong \angle 3$ therefore $\angle 1\cong \angle 2$

Hence, the congruency of angles is proved.

b.

To determine

To show that the distance travelled by the ball equals the distance $BH\text{'}$ .

Explanation of Solution

Given information:

$H\text{'}$ Is the image of H by reflection in the wall and $\overline{BH\text{'}}$ intersects the wall at P .

Proof:

  $\begin{array}{l}\Delta PXH\cong \Delta PXH\text{'}\\ PH=PH\text{'}\end{array}$

Distance travelled by the ball

$\begin{array}{l}=BP+PH\\ =BP+PH\text{'}\\ =BH\text{'}\end{array}$

Hence, it is proved that the distance travelled by the ball equals the distance $BH\text{'}$ .