Question

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, ______.

Options

• ABCD is a rhombus

• diagonals of ABCD are equal

• diagonals of ABCD are equal and perpendicular

• diagonals of ABCD are perpendicular

Answer 1

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, diagonals of ABCD are equal and perpendicular.

Explanation:

Given, ABCD is a quadrilateral and P, Q, R and S are the mid-points of sides of AB, BC, CD and DA, respectively.

Then, PQRS is a square.

∴ PQ = QR = RS = PS  ...(i)

And PR = SQ

But PR = BC and SQ = AB

∴ AB = BC

Thus, all the sides of quadrilateral ABCD are equal.

Hence, quadrilateral ABCD is either a square or a rhombus.

Now, in ΔADB, use mid-point theorem

SP || DB

And SP = `1/2` DB  ...(ii)

Similarly in ΔABC  ...(By mid-point theorem)

PQ || AC and PQ = `1/2` AC  ...(iii)

From equation (i), 

PS = PQ

⇒ `1/2` DB = `1/2` AC  ...[From equations (ii) and (iii)]

⇒ DB = AC

Thus, diagonals of ABCD are equal and therefore quadrilateral ABCD is a square not rhombus. So, diagonals of quadrilateral are also perpendicular.