Construction of angle

Question

Let $ABCD$ be the parallelogram and $P$ be the point lie on the perpendicular bisector of $AB$ such that $\angle{PBA}=\angle{PDA}$. What is the trick to construct this?

Answer 1

Say $\angle PBA =x$ then $\angle PAB =x$ so, since $\angle PDA =x$, circle around $APD$ touches $AB$. So, all you have to do is to draw this circle which is tangent to line $AB$ at $A$ and goes through $D$. This should be easy.

Notice that this point $P$ need not to exists and also that we colud have at most $2$ such points, which can be easly seen from the answer.