Question

$\angle BCD$ are $BO$ and $CO$ respectively, meets at $O$, then prove that
$\angle BOC={90}^o-\frac{\angle x}{2}$

Answer 1

$BO$ and $CO$ are angle bisector of $\angle EBC$ and $\angle DCB$

$\therefore \angle 1=\angle 2$ and $\angle 3=\angle 4$

$\angle EBC=\angle A+\angle C=\angle x+\angle z$ ($\because$ exterior angle) ...(1)

$\angle DCB=\angle B+\angle A=\angle y+\angle x$ ...(2)

$\angle EBC+\angle DCB=2\angle x+\angle y+\angle z$

$\angle 2+\angle 1+\angle 3+\angle 4=\angle x+{180}^o$

$\angle 2+\angle 3=\frac{\angle x}{2}+{90}^o$ ...(3)

In $\Delta BOC$,

$\angle 2+\angle 3+\angle BOC={180}^{\circ }$

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