Question 66
In triangle $ABC$, let $\angle BAC=60^\circ$. Let $E$ lie on line $AB$ with $B$ between $A$ and $E$ and $BE=BC$. Similarly let $F$ lie on line $AC$ with $C$ between $A$ and $F$ and $CF=BC$. The circumcircle of $ACE$ meets $EF$ at $K$. Prove that $K$ lies on the bisector of $\angle BAC$.