Question

In the given figure, the bisectors of $\angle ABC$ and $\angle BCA$ intersect each other at the point $O$. Prove that $\angle BOC=90+\frac{1}{2}\angle A$

Answer 1

In triangle $ABC$,

$\angle A+\angle B+\angle C={180}^{\circ }$ ..... (1)

$OB$ and $OC$ are bisectors of $\angle B$ and $\angle C$

So, $\angle B=2\angle OBC$

and $\angle C=2\angle OCB$

Now equation (1) can be written as,

$\angle A+2(\angle OBC+\angle OCB)={180}^{\circ }$ ..... (2)

In triangle $OBC$,

$\angle BOC+\angle OBC+\angle OCB={180}^{\circ }$

$\angle OBC+\angle OCB={180}^{\circ }-\angle BOC$..........(3)

From (2) and (3),

$\angle A+2({180}^{\circ }-\angle BOC)={180}^{\circ }$

$\angle A+{360}^{\circ }-2\angle BOC={180}^{\circ }$

$\angle A+{180}^{\circ }=2\angle BOC$

$\frac{1}{2}\angle A+{90}^{\circ }=\angle BOC$