Question 105
Let $\Omega$ be the circumcircle of a scalene triangle $ABC$ with $\angle ACB = 60^\circ$. The points $A'$ and $B'$ are chosen on the internal angle bisectors of the angles $BAC$ and $ABC$, respectively, so that $AB' \parallel BC$ and $BA' \parallel AC$. The line $A'B'$ meets $\Omega$ at points $D$ and $E$. Prove that the triangle $CDE$ is isosceles.