Question

Question 13
The circumcentre of the $\Delta ABC$ is O. Prove that $\angle OBC+\angle BAC={90}^{\circ }$.

Answer 1

Given a circle is circumscribed on a $\Delta ABC$ having centre O.

To prove that $\angle OBC+\angle BAC={90}^{\circ }$.
Construction : Join BO and CO.

Proof
Let $\angle OBC=\angle OCB=\theta$..........(i)
$In\Delta OBC,\angle OBC+\angle OCB+\angle COB={180}^{\circ }$ [by angle sum property of a triangle is ${180}^{\circ }$]
$\Rightarrow \angle BOC+\theta +\theta ={180}^{\circ }$

We know that in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\therefore \angle BOC=2\angle BAC$
$\Rightarrow \angle BAC=\frac{\angle BOC}{2}=\frac{{180}^{\circ }-2\theta }{2}={90}^{\circ }-\theta$
$\Rightarrow \angle BAC+\theta ={90}^{\circ }$
$\therefore \angle BAC+\angle OBC={90}^{\circ }[fromEq.(i\left)\right]$
Hence proved.