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Second grade math? Discrimination of parallelogram 2? A problem proved in many aspects.
Prove:

Method 1:

Because ABCD is a parallelogram

So AD=BC, angle ADF= angle EBC;; ; AB=CD, angle EBA= angle CDF

Because DF=BE

therefore

Triangle ADF is equal to triangle CBE, and triangle CDF is equal to triangle ABE.

So AF=CE, CF=AE

So the quadrilateral AECF is a parallelogram (an equilateral quadrilateral is a parallelogram)

Method 2:

Because the quadrilateral ABCD is a parallelogram, AB=CD, and AB is parallel to CD. So angle ABE= angle CDF. Because AB=CD, angle ABE= angle CDF, BE=DF. So the triangle is congruent triangles. So AE=CF, angle AEB= angle CFD. So 180 degree angle AEB= 180 degree angle CFD. So angle AEF= angle CFE. So AE is parallel to cf, because AE=CF and AE is parallel to CF, so the quadrilateral AECF is a parallelogram.