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 \(AE=BE\).