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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°

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