Chapter 5.1, Problem 5PSA

To determine

To Prove: $\overrightarrow{OB}\text{ does not bisect }\angle AOC$ .

Explanation of Solution

Given:

  

  $\overrightarrow{OB}\text{ is not an altitude}$

Concept Used:

Perpendicular line drawn from the centre of the circle bisects the chord.

Proof:

  $\begin{array}{l}\text{In }\Delta OAB\text{and }\Delta OCB\\ \overline{OA}\cong \overline{OC}\left(\text{Radius of circle}\right)\\ \overline{OB}\cong \overline{OB}\left(\text{Common side}\right)\\ \text{As here }\overrightarrow{OB}\text{ is not an altitude}\\ \Rightarrow m\angle OBA\ne m\angle OBC\ne {90}^{\circ }\\ \Rightarrow \angle OBA\cancel{\cong }\angle OBC\\ \Rightarrow \Delta OAB\cancel{\cong }\Delta OCB\\ \Rightarrow \angle AOB\cancel{\cong }\angle COB\\ \Rightarrow \overrightarrow{OB}\text{ will not bisect }\angle AOC\end{array}$