In convex pentagon $ABCDE$, the sides $AE,BC$ are parallel and $\angle ADE=\angle BDC$. The diagonals $AC$ and $BE$ intersect at $P$. Prove that $\angle EAD=\angle BDP$ and $\angle CBD=\angle ADP$.
Source: Baltic Way 1998
Tags: geometry proposed, geometry
In convex pentagon $ABCDE$, the sides $AE,BC$ are parallel and $\angle ADE=\angle BDC$. The diagonals $AC$ and $BE$ intersect at $P$. Prove that $\angle EAD=\angle BDP$ and $\angle CBD=\angle ADP$.