Question

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

Answer 1

In $△ABC,$

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

$⟹y+z={180}^{\circ }-x$

$⟹\frac{y+z}{2}={90}^{\circ }-\frac{x}{2}$...............(i)

$\angle CBO=\frac{1}{2}\angle CBE=\frac{1}{2}({180}^{\circ }-y)={90}^{\circ }-\frac{y}{2}$

$\angle BCO=\frac{1}{2}\angle BCD=\frac{1}{2}({180}^{\circ }-z)={90}^{\circ }-\frac{z}{2}$

In $△BOC,$

$\angle BOC+\angle CBO+\angle BCO={108}^{\circ }$

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

$⟹\angle BOC=\frac{y}{2}+\frac{z}{2}=\frac{y+z}{2}$

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

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

Hence proved.