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Question
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
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Solution
Let us join AC and BD.
In ΔABC,
P and Q are the mid-points of AB and BC respectively.
∴ 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
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.
∴ PS || QR and PS = QR ...(Opposite sides of the parallelogram) ...(3)
In ΔBCD, Q and R are the mid-points of side BC and CD respectively.
∴ QR || BD and QR = `1/2 BD` ...(Mid-point theorem) ...(4)
However, the diagonals of a rectangle are equal.
∴ AC = BD …(5)
By using equation (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus.
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