Question

In Figure, the sides AB and AC of $\Delta$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$={90}^o=\frac{1}{2}\angle$BAC.

Answer 1

REF. Imge.

BD is the bisector of $\angle CBE$

So, $\angle CBD=\angle EBO=\frac{1}{2}\angle CBE$

similarly

CO is the bisector of $\angle BCO$

So, $\angle BCO=\angle DCO=\frac{1}{2}\angle BCD$

$\angle CBE$ is the exterior angle of $△ABC$

Hence $\angle CBE=x+z\Rightarrow \frac{1}{2}\angle CBC=\frac{(x+z)}{2}$

$\angle CBO=\frac{1}{2}(x+z)$

$\angle BCD$ is the exterior angle of $△ABC$

$\angle BCD=x+y$

$\frac{1}{2}\angle BCD=\frac{1}{2}(x+y)$

$\angle BCO=\frac{1}{2}(x+y)$

in $△OBC,\angle BOC+\angle BCO+\angle CBO={180}^{\circ }$

$\angle BOC+\frac{1}{2}(x+y)+\frac{1}{2}(x+z)={180}^{\circ }$

$\angle BOC+\frac{1}{2}(x+y+x+z)={180}^{\circ }$

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

$\angle BOC+\frac{x}{2}+\frac{1}{2}\times {180}^{\circ }={180}^{\circ }$

$\angle BOC={90}^{\circ }-x/2$