Question

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

(a) ABCD is a rhombus.

(b) diagonals of ABCD are equal.

(c) diagonals of ABCD are equal and perpendicular.

(d) diagonals of ABCD are perpendicular.

Answer 1

Correct option is (c) diagonals of ABCD are equal and perpendicular.

In given figure,

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

Then, PQRS is a square.

∴ PQ = QR = RS = PS    ......(1)

and PR = SQ

But PR = BC and SQ = AB

∴ AB = BC

Thus, all sides of quadrilateral ABCD are equal.

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

Now, in △ADB,

By using Mid-point theorem,

SP ∣∣ DB; SP = \(\frac 12 \)DB    ......(2)

Similarly in △ABC,

PQ ∣∣ AC; PQ = \(\frac 12 \)​AC    ......(3)

From equation (1),

PS = PQ

From (2) and (3),

\(\frac 12 \)​DB = \(\frac 12 \)​AC

∴ DB = AC

Thus, diagonals of ABCD are equal and therefore quadrilateral ABCD is a square. So, diagonals of quadrilateral also perpendicular.