Chapter 5.1, Problem 40PPS

To determine

To prove: $\overline{AP}$ is the angle bisector of $\angle BAC$ .

Explanation of Solution

Given information:

The given triangle shown in figure (1).

  

Figure (1)

Th given relations are.

  $\angle BAC$ ; $P$ is in the interior of $\angle BAC$ , $PD=PE$ .

Proof:

Point $E$ is on the interior of $\angle BAC$ of $\Delta BAC$ .

Definition of congruence $\overline{PD}\cong \overline{PE}$ .

Definition of distance from a point to a line shows $\overline{PD}\perp \overline{AB}$ and $\overline{PE}\perp \overline{AC}$ .

  $\angle ADP$ and $\angle AEP$ are right angles by the definition of perpendicular lines.

  $\Delta ADP\cong \Delta AEP$ by HL congruence theorem.

  $\angle DAP\cong \angle EAP$

Which implies $\overline{AP}$ is the angle bisector of $\angle BAC$ .

From the above, $\overline{AP}$ is the angle bisector of $\angle BAC$ .

Hence, proved.