Question

Prove that $\angle BOC=90+\frac{1}{2}\angle BAC$

Answer 1

$\angle 1=\angle 2,\angle 4=\angle 5$

In $\Delta ABC$

$\angle A+\angle B+\angle C=180$

$\angle A+2\angle 2+2\angle 5=180$

$\angle 2+\angle 5=90-\frac{1}{2}\angle A$

Now in $\Delta BOC$

$\angle 2+\angle 5+\angle BOC=180$

$90-\frac{1}{2}\angle A+BOC=180$

$\angle BOC=90+\frac{1}{2}\angle A$

$\angle BOC=90+\frac{1}{2}\angle BAC$