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.