Berkeley Math Circle Monthly Contest 8

Solution:

Let the bisector of $\angle BAC$ intersect the circumcircle of $\triangle ACE$ at $K'$. The arcs, and hence the chords, $K'C$ and $K'E$ are equal; since $CB = BE$ is given, we have $\triangle K'BC \cong \triangle K'BE$ and so $K'$ is on the bisector of $\angle CBE$. This shows that $K'$ is the excenter of $\triangle ABC$ opposite $A$. By symmetry, we could have defined $K'$ as the intersection of the bisector of $\angle BAC$ and the circumcircle of $\triangle ABF$, and it would have been the same excenter.

To prove that $K' = K$, it remains to show that $F$, $K'$, and $E$ are collinear. Since $\angle CK'E = \angle BK'F = 120^\circ$ (by cyclic quads $AC K' E$ and $AB K' F$), it suffices to show that $\angle BK'C = 60^\circ$. But this follows from the properties of the excenter:
$$\begin{aligned} \angle BK'C & = 180 - \angle CBK' - \angle K'CB \\ & = 180 - \frac{1}{2} \angle CBE - \frac{1}{2} \angle FCB \\ & = 180 - \frac{360 - \angle ABC - \angle BCA}{2} \\ & = 180 - \frac{180 + \angle CAB}{2} \\ & = 180 - \frac{180 + 60}{2} = 60. \end{aligned}$$