Question

The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral formed by joining the midpoints of its sides is a rectangle.

Answer 1

Given: In quadrilateral ABCD, AC $\perp$ BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.

To prove: PQRS is a rectangle.

Proof:

In ΔABC, P and Q are mid-points of AB and BC, respectively.

∴ PQ || AC and PQ = $\frac{1}{2}$AC (Mid-point theorem) ...(1)

Similarly, in ΔACD,

So, R and S are mid-points of sides CD and AD, respectively.

∴ SR || AC and SR = $\frac{1}{2}$AC (Mid-point theorem) ...(2)

From (1) and (2), we get

PQ || SR and PQ = SR

But this is a pair of opposite sides of the quadrilateral PQRS,

So, PQRS is parallelogram.

Now, in ΔBCD, Q and R are mid-points of BC and CD, respectively.

∴ QR || BD and QR = $\frac{1}{2}$BD (Mid-point theorem) ...(3)

From (2) and (3), we get

SR || AC and QR || BD

But, AC ⊥ BD (Given)

∴ RS ⊥ QR

Hence, PQRS is a rectangle.