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.
Let us join AC and BD.
P and Q are the mid-points of AB and BC respectively.
∴ PQ || AC and PQ = 1/2AC (Mid-point theorem) ... (1)
Similarly in ΔADC,
SR || AC and SR = 1/2AC (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 parallelogram)... (3)
In ΔBCD, Q and R are the mid-points of side BC and CD respectively.
∴ QR || BD and QR =1/2BD (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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- The Mid-point Theorem