In the parallelogram ABCD, points E, F, K and M lie on its sides, as shown in the figure, with AE = CK, BF = DM. Prove that EFKM is a parallelogram
Consider the triangles AEM and CKF.
AE = CK (by condition of the problem)
∠A = ∠C (by the parallelogram property)
Because AD = BC (by the parallelogram property), and BF = DM (by the condition), then AM = CF.
Therefore, the triangles AEM and CKF are equal (by the first sign).
Therefore, EM = FK.
It is proved in a similar way that the triangles EBF and KDM are also equal, therefore EF = MK.
Those. opposite sides of a given quadrangle are equal. Accordingly, this quadrangle is a parallelogram (by the property of a parallelogram).
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