Question

If the bisectors BO and CO of $\angle CBE$ and $\angle BCD$ respectively meet at point O, then prove that $\angle BOC=90-\frac{1}{2}\angle A$

Answer 1

$\angle ABC=180-2x$
$\angle ACB=180-2y$
$\angle A+\angle ABC+\angle ACB=180$
$\angle A+(180-2x)+(180-2y)=180$
$\angle A=2x+2y-180$
$x+y=90+\frac{\angle A}{2}$
In $△BOC$
$x+y+\angle BOC=180$
$\angle BOC=180-(x+y)=180-(90+\frac{\angle A}{2})$
$=90-\frac{\angle A}{2}$