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Question

# Question 5 P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC⊥BD. Prove that PQRS is a square.

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Solution

## Given in quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC=BD and AC⊥BD. To prove PQRS is a square. Proof Now, in ΔADC, S are R are the mid-points of the sides AD and DC respectively, then by mid-point theorem, SR∥AC and PQ=12AC ……(i) In ΔABC,P and Q are the mid-points of AB and BC , then by mid- point theorem, PQ||AC and PQ=12AC …(ii) From Eqs. (i) and (ii) , PQ||SR and PQ=12AC …(iii) Similarly, in ΔABD, by mid-point theorem, SP||BD and SP=12BD=12AC [given, AC = BD] .....(iv) And ΔBCD, by mid-point theorem, RQ||BD and RQ=12BD=12AC [given, BD = AC] ...(v) From Eqs. (iv) and (v), SP=PQ=12AC From Eqs. (iii) and (iv) PQ = SR = SP= RQ Thus, all four sides are equal, Now, in quadrilateral OERF, OE || FR and OF || ER ∴∠EOF∠ERF=90∘ [∵AC⊥DB⇒=∠DOC=∠EOF=90∘ as opposite angles of a parallelogram] ∴∠QRS=90∘ Similarly ∠RQS=90∘ So, PQRS is a square. Hence proved.

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