show that the quadrilateral formed by joining the midpoint of the pair of adjacent side of a rhombus is a rectangle
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Let the rhombus have vertices A, B, C, D. Let the midpoints of the sides AB, BC, CD, DA be E, F, G, H.
If the diagonals of EFGH are equal and bisect each other then EFGH is a rectangle.
AB || HF || DC and AD || EG || BC.
So HF = AB = BC = EG.
ie diagonals HF and EG are equal in length.
HF bisects AD and BC. EG is parallel to AD and BC. Therefore HF bisects EG also.
By a similar argument, EG bisects HF.
So diagonals HE and EG (i) are equal in length, and (ii) bisect each other.
Therefore EFGH is a rectangle
If the diagonals of EFGH are equal and bisect each other then EFGH is a rectangle.
AB || HF || DC and AD || EG || BC.
So HF = AB = BC = EG.
ie diagonals HF and EG are equal in length.
HF bisects AD and BC. EG is parallel to AD and BC. Therefore HF bisects EG also.
By a similar argument, EG bisects HF.
So diagonals HE and EG (i) are equal in length, and (ii) bisect each other.
Therefore EFGH is a rectangle
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