Let $\triangle ABC$ be isosceles with $BC=AC$ and $\angle ACB=60^\circ$. Construct the circle with diameter $\overline{BC}$, meeting $\overline{AC},\overline{AB}$ at $D,E$. Let $F$ be the intersection of the diagonals of $BCDE$. What is $\angle BFC$?

1. $\angle CDB=\angle CEB=90^\circ$ (angles in a semicircle), and $\angle BFC=180^\circ-\angle A$.
2. With $\angle ACB=60^\circ$: base angles $\angle A=\angle B=60^\circ$ (equilateral!), so $\angle BFC=180^\circ-60^\circ=\boxed{120^\circ}$.