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
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