Question

In a $△ABC$ the bisector of $\angle ABC$ and $\angle BCA$ intersect at the point $0.$ Prove that $\angle BOC=$ ${90}^{\circ }+\angle \frac{A}{2}$

Answer 1

In $△ABC$, we have
$\angle A+\angle B+\angle C={180}^{\circ }($ sum of the angles of a triangle)
$\Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle B+\frac{1}{2}\angle C={90}^{\circ }$
$\Rightarrow \frac{1}{2}\angle A+\angle X+\angle y={90}^{\circ }\dots$(i)
Again in $△OBC$
$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$ (by angle sum property of a triangle)
$\Rightarrow \angle x+\angle y+\angle BOC={180}^{\circ }\dots$(ii)
Substituting $\angle x+\angle y={90}^{\circ }-\frac{1}{2}\angle A$ from (i) in eqn (ii), we get
${90}^{\circ }-\frac{1}{2}\angle A+\angle BOC={180}^{\circ }$
$\Rightarrow \angle BOC={180}^{\circ }-{90}^{\circ }+\frac{1}{2}\angle A$
$\Rightarrow \angle BOC={90}^{\circ }+\frac{1}{2}\angle A$
Hence proved.