Question

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

Answer 1

Given that.

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

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

So,

$\angle ABD=\angle CBD$ (By property)

To Prove:-

$\text{Seg}OD\perp \text{Seg}\text{AC}$

Proof:-

$\angle ABC={90}^0\text{(Angle}\text{inscribed}\text{in}\text{semicircle)}$

$\Rightarrow \angle ABD+\angle CBD={90}^0$

$\Rightarrow \angle ABD+\angle ABD={90}^0$

$\Rightarrow \angle 2ABD={45}^0$

$\Rightarrow \angle ABD={40}^0$

Also,

$\angle AOD=2\times \angle ABD(\text{Central}\text{Angle}\text{theorem)}$

$\Rightarrow \angle AOD=2\times {45}^0$

$SegOD\perp SegAC$

Hence, this is the answer.