Question

In the figure, the bisectors of $\angle$CBE and $\angle$ BCF intersect at G. Also $\overline{BE}||\overline{CF}$. Prove that $m\angle BGC={90}^o$.

Answer 1

$\Rightarrow \angle CBE+\angle BCF=180°$ ( Angle between two parallel lines)
$\frac{\angle CBE}{2}+\frac{\angle BCF}{2}=90°$ (From above)
$\angle CBG=\frac{\angle CBE}{2}$ and $\angle BCG=\frac{\angle BCF}{2}$
So, $\angle CBG+\angle BCG=90°$
Now, $\angle BGC+\angle CBG+\angle BCG=180°$
$\angle BGC+90°=180°$
$\angle BGC=90°$