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Prove that the quadrilateral formed by joining the mid-points of consecutive sides of a rectangle is a rhombus.

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#### Solution

Let us join AC and BD

In ΔABC

P and Q are the mid-point of AB and DC respectively

Therefore, PQ || AC and PQ = `(1)/(2)"AC"`(mid-point theorem) ...(1)

Similarly in ΔADC

SR || AC and SR = `(1)/(2)"AC"`(mid-point theorem) .....(2)

Clearly, PQ || SR and PQ = SR

As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.

Therefore, PS || QR and PS = QR (opposite sides of parallelogram)... (3)

Now, in ΔBCD, Q and R are mid points of side BC and CD respectively.

Therefore, QR || BD and QR = `(1)/(2)"BD"`(mid-point theorem)...(4)

But diagonals of a rectangle are equal

⇒ AC = BD ...(5)

Now, by using equation (1). (2), (3), (4), (5) we can say that

PQ = QR = SR = PS

So, PQRS is a rhombus.

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