Since ABCD is a parallelogram, the opposite sides will be parallel and equal,
[tex]\begin{gathered} AB=CD \\ BC=AD \end{gathered}[/tex]Consider that AC acts as a transversal to the parallel lines AB and CD, so we can write,
[tex]\begin{gathered} \angle CAD=\angle ACB\text{ (Alternate Interior Angles)} \\ BC=AD\text{ (Opposite sides of parallelogram)} \\ \angle ADB=\angle CBD\text{ (Alternate Interior Angles)} \end{gathered}[/tex]So by the ASA criteria, the triangle AED is congruent to the triangle CEB,
Then the corresponding parts of the triangles will be equal,
[tex]\begin{gathered} AE=CE \\ BE=DE \end{gathered}[/tex]Hence Proved.