Q. 24.1( 374 Votes )

# ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Answer :

The figure is given below:

**To Prove: PQRS is a rectangle.For that, we need to prove that, PQ = RS, PS = QR andalso, that angles of PQRS are all right angles.**

**Proof:**

We start by proving the equality of sides of quadrilateral PQRS

We start by proving the equality of sides of quadrilateral PQRS

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

The **Midpoint Theorem** states that the segment joining two sides of a triangle at the midpoints

of those sides is parallel to the third side and is half the length of the third side.

∴ From the mid-point theorem,

PQ || AC &

PQ = ½ AC ..(1)

Also, R and S are the mid-points of CD and AD respectively

∴ From the mid-point theorem,

RS || AC

& RS = ½ AC ..(2)

Therefore,

From equations (1) and (2), we obtain

PQ || RS and PQ = RS

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other

So it is a parallelogram.

Let the diagonals of rhombus ABCD intersect each other at point O

In quadrilateral OMQN,

MQ || ON (Because PQ || AC)

QN|| OM (Because QR || BD)

Therefore,

OMQN is a parallelogram

∠MQN = ∠NOM

∠PQR = ∠NOM

However, NOM = 90^{0} (Diagonals of a rhombus are perpendicular to each other)

So, PQR = 90^{0}

Clearly, PQRS is a parallelogram having one of its interior angles as 90^{0}

Hence, PQRS is a rectangle**Hence, Proved.**

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