Question

Question 3
ABCD is a rectangle and P, Q, R, 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 $\Delta$ ABC,
P and Q are the mid-points of AB and BC respectively.
PQ || AC and PQ = $\frac{1}{2}$AC ( Mid - Point theorem) ….(1)
Similarly in $\Delta$ ADC,
SR || AC and SR = $\frac{1}{2}$ AC ( Mid - point theorem ) ….(2)
Clearly, From (1) and (2) 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 parallelogram)…(3)
In $\Delta$ BCD, Q and R are the mid-points of side BC and CD respectively.
$\therefore$ QR || BD and QR = $\frac{1}{2}$ ( Mid - point theorem ) ….(4)
However, the diagonals of a rectangle are equal.
$\therefore$ AC = BD ....(5)
By using equation (1), (2) , (3) and (4) and (5) , we obtain
PQ=QR=SR=PS
Therefore, PQRS is a rhombus.