It is known that in ▱ ABCD, M is the midpoint of edge ad, and BM = cm

It is proved that the ∵ quadrilateral ABCD is a parallelogram, ∵ AB = CD, ab ∥ CD, ∵ a + ∠ d = 180 ° in △ ABM and △ DCM, am = DMAB = dcbm = cm, ≌ ABM ≌ DCM, ≌ a = ∠ d = 90 ° we can conclude that the parallelogram ABCD is a rectangle

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