External Bisectors of Triangle ABC

Question

The exterior angle bisectors of $\angle B$ and $\angle C$ intersect on point $O$. $\angle BOC=70°$.

Find $\angle OAC$.

Answer 1

We know that exterior angle of $\angle b$ = $2\beta$ = $\angle A + 180 - \angle c$ Suppose exterior angle of $\angle c$ is $2\alpha$, so $$2\beta = A + 180 - 2\alpha$$, but $\alpha + \beta = 180 - 70 = 110$,so $$A = 220 - 180 = 40$$ That's was part one, now part two. Let's prove that AO is bisector. Let's take the heights on BC , (AB) and (AC), where (AB) mean line AB. So as BO and CO are bisectors, then it's easy to prove that O - is center of some circle , which intersect BC, (AC), (AB) , so O is lying on AO - bisector, so $\angle OAC = 20$