Chapter 5.1, Problem 10PSB

To determine

To Prove: $\overline{AB}\cancel{\cong }\overline{AC}$ .

Explanation of Solution

Given:

  

  $\overline{PA}\perp \overline{AB},\overline{PA}\perp \overline{AC},\angle B\cancel{\cong }\angle C$

Concept Used:

Two line segments are said to be congruent if they are of the same measure.

Proof:

Here, we have

  $\overline{PA}\perp \overline{AB},\overline{PA}\perp \overline{AC},\angle B\cancel{\cong }\angle C$

  $\begin{array}{l}\text{In }\Delta PAB\text{and }\Delta PAC\\ \Rightarrow \angle PAB=\angle PAC={90}^{\circ }\\ \Rightarrow m\angle PAB=m\angle PAC\\ \overline{PA}\cong \overline{PA}\left(\because \text{Common side}\right)\\ \angle B\cancel{\cong }\angle C\left(\text{Given}\right)\\ \Rightarrow \Delta PAB\cancel{\cong }\Delta PAC\\ \therefore \overline{AB}\cancel{\cong }\overline{AC}\end{array}$