ABCD is a parallelogram. AE = CK, BF = DM. Prove that EFKM is a parallelogram.

In a parallelogram, the lengths of the opposite sides are equal and the opposite angles are also equal.

By condition, AE = CK, then BE = DK, since AB = CD. BF = DM, angle ABC = ADC, then the triangles EBF and MDK are equal on two sides and the angle between them, which means EF = MK.

Similarly, in triangles AEM and KCF the angle BAD = BCD, AE = CK, AM = CF, then the triangles are equal, which means EM = FK.

In the quadrangle EFKM, the opposite sides are pairwise equal, hence the quadrilateral is a parallelogram, one hundred and it was required to prove.