Chapter 3.3, Problem 20PSB

To determine

To prove : $\overrightarrow{OQ}$ bisects $\angle PQR$ .

Explanation of Solution

Given information : In this question, we have given that center of circle is O and $\overline{PQ}\cong \overline{QR}$

Formula used: SSS Congruency rule.

Construction: Join $\overline{OP}$ and $\overline{OR}$

Proof: As, O is the center of the circle and P and R are points on the circle,

  $\therefore \overline{OP}$ and $\overline{OR}$ are radius

Now, In $\Delta OQP$ and $\Delta OQR$

  $\overline{OP}\cong \overline{OR}$ (Radius of circle are equal)

  $\overline{PQ}\cong \overline{QR}$ (Given)

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

  $\therefore \Delta OQP\cong \Delta OQR$ (BY SSS Congruency rule)

  $\Rightarrow \angle OQP\cong \angle OQR$ (BY CPCT)

  $\therefore$ $\overrightarrow{OQ}$ bisects $\angle PQR$ .