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Question
The diagonals PR and QS of a quadrilateral PQRS are perpendicular to each other. A, B, C and D are mid-point of PQ, QR, RS and SP respectively. Prove that ABCD is a rectangle.
Solution
In ΔPQS, A and D are mid-points of sides QP and PS respectively.
Therefore, AD || QS and AD = `(1)/(2)"QS"` ..........(i)
In ΔQRS
B and C are the mid-points of QR and RS respectively
Therefore, BC || QS and BC = `(1)/(2)"QS"` ..........(ii)
From equations (i) and (ii),
AD || BC and AD = BC
As in quadrilateral ABCD one pair of opposite sides are equal and parallel to each other, so it is parallelogram.
The diagonals of quadrilateral PQRS intersect each other at point O.
Now in quadrilateral OMDN
ND || OM ...( AD || QS)
DM || ON ...(DC || PR)
So, OMDN is parallelogram
∠MDN = ∠NOM
∠ADC =∠NOM
But, ∠NOM = 90° ...(doagonals are perpendicular to each other)
⇒ ∠ADC = 90°
Clearly ABCD is a parallelogram having one of its interior angle as 90°.
Hence, ABCD is rectangle.
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