Let ABCD be a rectangle and let P, Q, R, S be the mid-points of AB, BC, CD, DA respectively. Prove that PQRS is a rhombus.

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Solution

Given: AP = BP, BQ = CQ, DR = CR and AS = SD

We know that each angle of a rectangle measures 90°. Therefore, triangles APS, PBQ, QCP and SRD are congruent to each other.

∴ SP = PQ = QR = SR

We know that in any triangle, the line joining the midpoints of two sides is parallel to the third side.

∴ SP || BD and RQ || BD

∴ SP || RQ

Similarly, it can be shown that SR || PQ

Hence, PQRS is a rhombus.

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