And BF=BC, so BF=BC=EF.

Because of AD‖FE, quadrilateral BCEF is a parallelogram. BF=BC, so the quadrilateral BCEF is a diamond.

(2) BE=CF from (1)

And ab = BC = CD, BF = BC, AD‖FE.

Therefore, the quadrangles ABEF and CDEF are parallelograms, so AF=BE and CF=DE. And AC=BD.

In triangle ACF and BDE, AF=BE, AC=BD, CF=DE.

So the triangle δACFδBDE