Chapter 3, Problem 8CR

To determine

To prove : $\overrightarrow{PQ}$ bisects $\angle RPS$ .

Explanation of Solution

Given information : Here, we have given a circle with center Q and $\overline{RP}\cong \overline{PS}$

Construction: Join $\overline{QR}$ and $\overline{QS}$ .

Proof: In $\Delta QPR$ and $\Delta QPS$ ,

  $\overline{RP}\cong \overline{PS}$ (Given)

  $\overline{QR}\cong \overline{QS}$ (Radius of circle)

  $\overline{QP}\cong \overline{QP}$ (Common)

  $\Rightarrow \Delta QPR\cong \Delta QPS$ (By SSS Congruency Rule)

  $\Rightarrow \angle QPR\cong \angle QPS$ (By Corresponding Part of Congruent Triangles)

  $\therefore$

  $\overrightarrow{PQ}$ bisects $\angle RPS$