Let ABCD be a trapezium with AB and CD being parallel. Let P and Q be points lying on AD and BC respectively, such that $\angle$APB=$\angle$CPD and $\angle$AQB=$\angle$CQD. Let E be the point of intersection of diagonals AC and BD.
Prove that point E lies on the perpendicular bisector of PQ.
My idea for solving this was to prove that the triangle $\triangle$PEQ is isosceles, which would give us the thesis instantly. I've also managed to prove that angles $\angle$APB=$\angle$CPD=$\angle$AQB=$\angle$CQD, which also means that points A, B, P, Q lie on one circle, and points C, D, P, Q lie on another. I've tried to apply Stewart's Theorem but it didn't get me far. For example, if you draw a line parallel to PC going through point B we obtain an isosceles triangle with cevian AB with third vertex being K. By Stewart's Theorem and simmilar triangles $\triangle$CPD and $\triangle$BKA we have
$AB^2$=$BP^2$-AP$\times$DP$\frac{AB}{CD}$.
Like so we obtain:
$AB^2$=$AQ^2$-BQ$\times$CQ$\frac{AB}{CD}$
$CD^2$=$CP^2$-AP$\times$DP$\frac{CD}{AB}$
$CD^2$=$DQ^2$-BQ$\times$CQ$\frac{CD}{AB}$.
I've also tried applying analytical geometry but that got me nowhere.
After that I've encountered a wall. Would really appreciate help.