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Question
The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the midpoints of quadrilateral ABCD is a rectangle.
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Solution
The figure is as below:
Let ABCD be a quadrilateral where p, Q, R, S are the midpoint of sides AB, BC, CD, DA respectively. Diagonals AC and BD intersect at right angles at point O. We need to show that PQRS is a rectangle
Proof:
In 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,
∴ ∠AOD = ∠PXO = 90° ....(Corresponding angles)
Again, BD || RQ,
∴ ∠PXO = ∠RQX = 90° ....(Corresponding angles)
Similarly, ∠QRS = ∠RSP = SPQ = 90°
Therefore, PQRD is a rectangle.
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