Given an equilateral Δ ABC, D, E and F are the mid-points of AB, BC and AC respectively. Then the quadrilateral BEFD is exactly a :
AS, ΔABC is equilateral and D, E and F are the mid-points of AB, BC and AC respectively. BE=DB and EF=DF.
Since , ∠B = 60. It cannot be a square.
So , it is a rhombus.
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