Question

In figure the sides $AB$ and $AC$ of $\Delta ABC$ are produced to point $E$ and $D$ repectively. If bisector $BO$ and $CO$ of $\angle CBE$ and $\angle BCD$ respectively meet a point $O$, then $\angle BOC={90}^{\circ }-\frac{1}{2}\angle BAC$

Answer 1

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

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

Similarly,$CO$ is the bisector of $\angle BCD$.

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

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

Hence, $\angle CBE$$=x+z$

(External angle sum of two interior opp angles)

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

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

Similarly, $\angle BCD$ is the exterior angle of $\Delta ABC$.

Hence, $\angle BCD$$=x+y$

(External angle is sum of two interior opposite angles)

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

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

In $\Delta OBC$,

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

(angle sum property of triangles)

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

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

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

In $\Delta ABC$,

$x+y+z={180}^o$ ………..$\left(2\right)$ (Angle sum property of triangle)

Putting $\left(2\right)$ in $\left(1\right)$

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

$\angle BOC+\frac{1}{2}(x+180)=180$

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

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

$\angle BOC=180-90=\frac{x}{2}$

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

$\angle BOC={90}^{\circ }-\frac{1}{2}\times \angle BAC$

$\therefore$ Hence proved.