Question

A quadrilateral ABCD is drawn in which the mid points of sides AB, BC, CD, and AD are P, Q, R, and S respectively.

If quadrilateral ABCD is a rectangle, show that the quadrilateral PQRS is a rhombus.

Answer 1

As ABCD is a rectangle, the diagonals AC and BD bisect each other and are equal in length.

By mid point theorem,

PS = QR = $\frac{1}{2}$BD --------------------------- I

PQ = RS = $\frac{1}{2}$AC ---------------------------- II

But, BD = AC

Therefore, PQ = QR = RS = PS => Sides are equal

So, PQRS is either a rhombus or a square.

Also, $△$AOB ≅ $△$SRQ By SSS criteria

Thus, ∠AOB = ∠SRQ (by CPCTE)

We know that, diagonals of a rectangle do not intersect at 90°.

Therefore, ∠SRQ ≠ 90°

Hence, PQRS is a rhombus.