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प्रश्न
The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is rectangle.
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उत्तर
The figure is shown below
Let ABCD be a quadrilateral where P, Q, R, S are the midpoint of AB, BC, CD, DA. Diagonal AC and BD intersect at a right angle at point O. We need to show that PQRS is a rectangle
Proof:
From and ΔABC and ΔADC
2PQ = AC and PQ || AC …..(1)
2RS = AC and RS || AC …..(2)
From (1) and (2) we get,
PQ = RS and PQ || RS
Similarly, we can show that PS=RQ and PS || RQ
Therefore PQRS is a parallelogram.
Now PQ || AC, therefore ∠AOD = ∠PXO = 90° ...[ Corresponding angel ]
Again BD || RQ, therefore ∠PXO = ∠RQX = 90° ...[ Corresponding angel]
Similarly ∠QRS = ∠RSP = ∠SPQ = 90°
Therefore PQRS is a rectangle.
Hence proved.
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P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.
