Question

In fig. sides AB and AC of triangle ABC are produce to E and D respectively. If angle bisectors BO and CO of $\angle CBE$ and $\angle BCD$ meet each other at point O , then prove that :
$\angle BOC={90}^{\circ }-\frac{\angle x}{2}$

Answer 1

$x+y+z={180}^{\circ }$ (sum of the angles of a triangle)

$\angle CBE=x+z$ (sum of the interior angles=opposite exterior angles)

$\angle BCD=x+y$ (sum of the interior angles=opposite exterior angles)

Also,

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

Similarly,

$\angle ACB+\angle BCD={180}^{\circ }\Rightarrow x={180}^{\circ }-\angle BCD\therefore \angle CBE=x+z={180}^{\circ }-\angle BCD+x\Rightarrow \angle CBE+\angle BCD={180}^{\circ }+x$

$\Rightarrow 2\angle CBO+2\angle BCO={180}^{\circ }+x$ ($BO$ & $CO$ bisects the angles)

$\Rightarrow \angle CBO+\angle BCO={90}^{\circ }+\frac{x}{2}\Rightarrow {180}^{\circ }-\angle BOC={90}^{\circ }+\frac{x}{2}\Rightarrow \angle BOC={90}^{\circ }-\frac{x}{2}$

Hence proved.