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.

Answer 1

Let us join AC and BD.

In ∆ABC,

P and Q are the mid-points of AB and BC respectively

∠PQ || AC and $\frac{1}{2}$ PQ = AC (Mid-point theorem) ... (1)

Similarly in

∆ADC,

SR || AC and SR = $\frac{1}{2}$ 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,

Hence, it is a parallelogram.

∠PS || QR and PS = QR (Opposite sides of parallelogram)... (3)

In ∆BCD, Q and R are the mid-points of side BC and CD respectively.

∠QR || BD and QR = $\frac{1}{2}$ 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.