Let: $\angle BAC=\angle BCA=\alpha$ By cyclic we get: In $ABCD$: $\angle ADB=\angle BDC=\alpha$ In $ABCE$ : $\angle BCA=\angle BEA=\alpha$ $\implies PQED$ is cyclic. $\implies \angle PQE=\angle PDE=\angle CAE$ $\implies PQ \parallel AC$.$\blacksquare$

A quick angle chase yields $$\angle ADB = \angle ACB = \angle BAC = \angle BDC = \angle AEB,$$so $PEDQ$ is cyclic. Then, $$\angle PQE = \angle CDE = \angle CAE,$$which implies $PQ \parallel AC$.