Question

The sides AB and AC of ABC are produced to point P and Q respectively . If the bisectors of $\angle PBC$ and $\angle QCB$ meet at point O , then prove that $\angle BOC$ = 90 - $\frac{1}{2}$ $\angle A$ .

Answer 1

Since $\angle ABC$ and $\angle CBP$ form a linear pair.

$\angle ABC+\angle CBP={180}^o$

$\angle B+2\angle 1={180}^o$ [BO is the bisector of $\angle$ CBP]

$2\angle 1={180}^o-\angle B$

$\angle 1={90}^o-\frac{1}{2}\angle B$ .....(1)

Again, $\angle ACB$ and $\angle QCB$ form a linear pair.

Therefore,

$\angle ACB+\angle QCB={180}^o$

$\angle C+2\angle 2={180}^o$

$2\angle 2={180}^o-\angle C$

$\angle 2={90}^o-\frac{1}{2}\angle C$ ...(2)

In $△BOC$, we have

$\angle 1+\angle 2+\angle BOC={180}^o$

$90-\frac{1}{2}\angle B+{90}^o-\frac{1}{2}\angle C+\angle BOC={180}^o$

${180}^o-\frac{1}{2}(\angle B+\angle C)+\angle BOC={180}^o$

$\angle BOC=\frac{1}{2}(\angle B+\angle C)$

$\angle BOC=\frac{1}{2}({180}^o-\angle A)$

$\angle BOC={90}^o-\frac{1}{2}\angle A$