Question

In Fig. the sides $AB$ and $AC$ of $△ABC$ are produced to points E and D respectively. If bisectors BO and CO of $\angle CBE$ and $\angle BCD$ respectively meet at point O, then prove that $\angle BOC={90}^0-\frac{1}{2}\angle BAC.$

Answer 1

$\angle CBE={180}^{\circ }-y$

$\therefore \angle CBO=\frac{\angle CBE}{2}={90}^{\circ }-\frac{y}{2}$

Similarily,

$\angle BCO={90}^{\circ }-\frac{z}{2}$

$\angle BAC={180}^{\circ }-(\angle ABC+\angle ACB)$

$x={180}^{\circ }-(y+z)$

$y+z={180}^{\circ }-x$

$\angle BOC={180}^{\circ }-(\angle CBO+\angle BCO)$

$\angle BOC={180}^{\circ }-({90}^{\circ }-(\frac{y}{2})+{90}^{\circ }-\frac{z}{2})$

$\angle BOC={180}^{\circ }-({180}^{\circ }-(\frac{y+z}{2})$

$\angle BOC=\frac{y+z}{2}={90}^{\circ }-\frac{x}{2}$