6.4 Triangles: Angle Sum & Exterior Angles

In $\Delta ABC$, $\angle A + \angle B + \angle C = 180^\circ$. So $\angle B + \angle C = 180^\circ - \angle A$.

Divide by 2: $\frac{1}{2}\angle B + \frac{1}{2}\angle C = 90^\circ - \frac{1}{2}\angle A$.

In $\Delta BOC$, $\angle OBC + \angle OCB + \angle BOC = 180^\circ$.

Since OB and OC are bisectors, $\angle OBC = \frac{1}{2}\angle B$ and $\angle OCB = \frac{1}{2}\angle C$.

Substitute: $(\frac{1}{2}\angle B + \frac{1}{2}\angle C) + \angle BOC = 180^\circ$.

Substitute first equation: $(90^\circ - \frac{1}{2}\angle A) + \angle BOC = 180^\circ$.

Solve for $\angle BOC$: $\angle BOC = 180^\circ - 90^\circ + \frac{1}{2}\angle A$.

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