Question

In the figure, $seg$ $AC$ is a diameter of the circle with centre $O.$ Bisector of $\angle ABC$ intersects the circle at $D.$ Prove that $,seg$ $OD\perp seg$ $AC$

Answer 1

$AC$ is the diameter of the circle with the center $C$.

$BD$ is the angle bisector of angle $ABC$.

That is , $\angle ABD=\angle CBD$ (by the property of angle bisector)

To prove : $segOD\perp segAC$

Proof ; Since, $\angle ABC={90}^0$ (Angle inscribed in semicircle )

$\begin{array}{c}\Rightarrow \angle ABD+\angle CBD={90}^0\\ \Rightarrow \angle ABD+\angle ABD={90}^0\\ \Rightarrow 2\times \angle ABD={90}^0\\ \Rightarrow \angle ABD={45}^0\\ Also,\angle AOD=2\times \angle ABD\left(centralangletheorem\right)\\ \Rightarrow \angle AOD=2\times {45}^0\\ \Rightarrow \angle AOD={90}^0\\ \Rightarrow OD\perp AC\end{array}$

Hence, Proved