# 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.

In ΔABC,

P and Q are the mid-points of AB and BC respectively To Prove: PQRS is a rhombus.
Proof:
The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

PQ || AC and PQ = AC (Mid-point theorem) (1)

SR || AC and SR = 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, 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 = 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

Now, as all sides of the rhombus are equal.

Hence, PQRS is a rhombus

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