Draw parallelogram ABCD. Mark E at the midpoint of AB. Join CE. Let ∠BCE = ∠ECD. Draw EF parallel to BC meeting CD at F.
As EBCF is a parallelogram, ∠BEC = ∠ ECF = x.
Hence, in ΔEBC, EB = BC.
Hence, EB = BC = CF = FD = EF = AE = AD
Now, quadrilaterals AEFD and EBCF are Rhombuses as opposite sides are parallel and the sides are equal.
Their diagonals are perpendicular.
Hence, EC ⊥ BF.
We know that ED || BF
hence, EC ⊥ ED. Hence, ∠ DEC = 90°