Let $ \vartriangle ABC $ be a triangle with $ \angle ACB = 60º $. Let $ E $ be a point inside $ \overline {AC} $ such that $ CE <BC $. Let $ D $ over $ \overline {BC} $ such that $$ \frac {AE} {BD} = \frac {BC} {CE} -1 .$$Let us call $ P $ the intersection of $ \overline {AD} $ with $ \overline {BE} $ and $ Q $ the other point of intersection of the circumcircles of the triangles $ AEP $ and $ BDP $. Prove that $QE \parallel BC $.