Chapter 4.7, Problem 17WE

To determine

To prove: if a point is equidistant from the sides of an angle, then the point lies on the bisector of the angle.

Explanation of Solution

Given information:

Given figure,

Given,

$\overline{PX}\perp \overrightarrow{BA}$ ; $\overline{PY}\perp \overrightarrow{BC}$

  $PX=PY$

Calculation:

Key steps of proof are as follows:

  $\angle PXB=\angle PYB\left(\overline{PX}\perp \overline{BA};\overline{PY}\perp \overline{BC}\right)$

  $PX=PY$

  $BP=BP\left(\text{Common}\text{side}\right)$

  $\Delta PXB\cong \Delta PYB\left(\text{By}\text{AAS}\text{Postulate}\right)$

  $\angle XBP=\angle YBP\left(\text{Corr}\text{.}\text{parts}\text{of}\cong \Delta s\right)$

$\overrightarrow{BZ}$ Bisects $\angle ABC$ . $\left(\angle XBP=\angle YBP\right)$