Problem #PRU-5140

Problem

In the quadrilateral $ABCD$ the diagonals $AC$ and $BD$ intersect at the point $E$. It is known that the perimeter of the triangle $ABC$ is equal to the perimeter of the triangle $ABD$, and the perimeter of the triangle $ACD$ equals the perimeter of the triangle $BCD$.

Prove that $AE=BE$.