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ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.

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We know that the diagonals of a rhombus are perpendicular bisector of each other

In ΔBDE A O , and are mid points of BE and BD respectively

 In ΔCFA B O , and are mid points of AF and AC respectively

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