Finding the angle measure of a triangle inscribed in a circle 
I'm working on this question for an online math class and I don't know how to start on this; I've been stuck for awhile. Can someone help me out?
 A: Let $X$ be the intersection of $BE$ and $CF$, and let $M$ be the midpoint of $BC$. By moving ratios in the diagram, we’ll prove $\angle A=\boxed{60^\circ}$.
First, notice that $$\bigtriangleup ABC\sim\bigtriangleup AEF,$$ and that furthermore, these two triangles are related by a reflection on $\angle ABC$ and a scaling. This implies that $$\angle BAK=\angle CAO.$$ However, we also have $$\angle BAK=\angle CAX$$ – this is a very well-known lemma that amounts to angle chasing. Therefore, $A$, $K$, $O$ are collinear, and $$AO=2AK.$$ Once again due to our similarity, we can calculate $$\frac{AK}{AX}=\frac{AE}{AB}\Rightarrow$$ $$BO=4OM\cdot\frac{AE}{AB},\label{1}\tag{1}$$ using that $AO=BO$, and another well-known lemma that states that $AX=2OM$ (this lemma can be proven by constructing the midpoint $N$ of $AC$ and noticing that $\bigtriangleup AXB\sim\bigtriangleup MON$; this lemma also bears relation to the Euler Line). Furthermore, since $$\bigtriangleup BOM\sim\bigtriangleup BAE$$ by AA, $$\frac{AE}{AB}=\frac{OM}{OB}\label{2}\tag{2}$$ Therefore, by $(\ref{1})$ and $(\ref{2})$, $$BO^2=(2OM)^2\Rightarrow$$ $$BO=2OM.$$ Therefore, $\bigtriangleup BOM$ is a $30^\circ-60^\circ-90^\circ$, and $$60^\circ=\angle BOM=\frac12\angle BOC=\angle BAC.$$ $\blacksquare$
