Question

In given fig. sides $AB$ and $AC$ of $\Delta ABC$ are produced 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

$BO$ Is the bisector of $\angle CBE$

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

Similarly,

$CO$ is bisector of $\angle BCD$

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

$\angle CBE$ is the exterior angle of $\Delta ABC,$

So,

$\angle CBE=x+z$ (External angle is sum of two interior opposite angle )

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

Similarly,

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

Hence,

$\angle BCD=x+y$

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

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

In $\Delta OBC$

$\angle BOC+\angle BCO+\angle CBO={180}^0$

$\angle BOC+\frac{1}{2}(x+y)+\frac{1}{2}(x+z)={180}^0$

$\angle BOC+\frac{1}{2}(x+y+x+z)={180}^0$

In $△ABC$

$x+y+z={180}^0$

Hence,

$\angle BOC+(x+y+x+z)={180}^0$ $

$\angle BOC+\frac{x}{2}+{90}^0={180}^0$

$\angle BOC={90}^0-\frac{x}{2}$