Given triangle $ ABC$. Point $ O$ is the center of the excircle touching the side $ BC$. Point $ O_1$ is the reflection of $ O$ in $ BC$. Determine angle $ A$ if $ O_1$ lies on the circumcircle of $ ABC$.
Is this really trivial or am I wrong?
We have that $ < OBC = 90 - <\frac {B}{2}$, which implies that $ < BO_{1}O = <\frac {B}{2}$.Similarly, $ < CO_{1}B = \frac {C}{2}$
But $ O_{1}ABC$ is cyclic, then $ < BAC = < BO_{1}C$.
Hence, $ \frac {3}{2}(180 - < A) = 180$ and then $ < A = 60$.
Probably I am making some mistakes or misunderstanding the problem. I didn't get $ 60^{\circ}$ like msecco.
I got $ \angle A = 36^{\circ}$.
We have $ \angle OBC = 90^{\circ} - \frac B2$ and $ \angle OCB = 90^{\circ} - \frac C2$. Which implies $ \angle BOC = \frac B2 + \frac C2 = 90^{\circ} - \frac A2 = 2A \iff A = 36^{\circ}$.