Question

The midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC = BD and AC ⊥ BD then prove that the quadrilateral formed is a square.

Answer 1

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

To prove: PQRS is a square.

Construction: Join AC and BD.

Proof:

In ΔABC,

$\because$ P and Q are mid-points of AB and BC, respectively.

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

Similarly, in ΔACD,

$\because$ R and S are mid-points of sides CD and AD, respectively.

$\therefore$ 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 a pair of opposite sides of the quadrilateral PQRS.

So, PQRS is parallelogram.

Now, in ΔBCD,

$\because$ Q and R are mid-points of sides BC and CD, respectively.

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

From (2) and (3), we get

RS || AC and QR || BD

But, AC ⊥ BD (Given)

∴ RS ⊥ QR

But this a pair of adjacent sides of the parallelogram PQRS.

So, PQRS is a rectangle.

Again, AC = BD (Given)

$\Rightarrow$ $\frac{1}{2}$AC = $\frac{1}{2}$BD

$\Rightarrow$ RS = QR [From (2) and (3)]

But this a pair of adjacent sides of the rectangle PQRS.

Hence, PQRS is a square.