In the given figure,AB∥DC
Hence,∠CDB=∠DBA (Alternate angles)
Now in triangles DEC and BEA
∠DEC=∠BEA (Vertically opposite angles)
∠CDB=∠DBA (Proved above)
EC=EA (E is midpoint of AC)
Hence, by AAS criterion triangle DEC ≅ triangle BEA
Then by c.p.c.t. DE=BE
Therefore E is the midpoint of BD