Question

The circumference of a $\Delta ABC$ is O. prove that $\angle OBC+\angle BAC={90}^{\circ }$.

Answer 1

$ABC$ is a triangle and $O$ is the circumcentre.

Refer image,

Draw $OD\perp BC$, Join $OB$ and $OC$

In right $\Delta OBD$ and right $\Delta OCD$, we have,

hyp. $OB=hyp.OC$

[Radii of the same circle]

$OD=OD$

[Common side]

$\therefore \Delta OBD\cong \Delta OCD$ [By RHS cong. Role]

$\therefore \angle 1=\angle 2$ and $\angle 3=\angle 4$ [By C.P.C.T]

Now, $\angle BOC=2\angle 1$ and $\angle BOC=2\angle A$

$\therefore 2\angle 1=2\angle A\Rightarrow \angle 1=\angle A$

$\therefore \angle A=\angle 2$......(1) $[\because \angle 1=\angle 2]$

$\Rightarrow \angle A+\angle 4=\angle 2+\angle 4$ [Adding $\angle 4$ to both sides]

$\Rightarrow \angle A+\angle 3={90}^0[\because \angle 2$ +$\angle 4={90}^{0}and\angle 4=\angle 3]$

$\Rightarrow BOC=\angle A={90}^0$

Hence, proved.