Option : (B)
Let ABCD is a rectangle such as AB = CD and BC = DA
P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively
Construction :
Join AC and BD
In ΔABC,
P and Q are the mid-points of AB and BC respectively
Therefore,
PQ ‖ AC and PQ = \(\frac{1}{2}\)AC (Mid-point theorem) ...(i)
Similarly,
In ΔADC,
SR ‖ AC and SR = \(\frac{1}{2}\)AC (Mid-point theorem) ..(ii)
Clearly, from (i) and (ii)
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.
Therefore,
PS ‖ QR and PS = QR (Opposite sides of a parallelogram) ...(iii)
In ΔBCD,
Q and R are the mid-points of side BC and CD respectively
Therefore,
QR ‖ BD and QR = \(\frac{1}{2}\)BD(Mid-point theorem) ...(iv)
However,
The diagonals of a rectangle are equal
Therefore,
AC = BD ...(v)
Now,
By using equation (i), (ii), (iii), (iv), and (v), we obtain
PQ = QR = SR = PS
Therefore,
PQRS is a rhombus.