Question

ABCD is a rhombus P Q R and S are the midpoints of the sides ab BC CD and AD respectively show that pqrs is a rectangle​

Answer 1

Answer:

Step-by-step explanation:

U have to remember the main thing and use is the mid point theorem....I used it and elongated it only once....But u have to do it everytime....Hope that this clarified ur doubt upto the maximum extent....

Answer 2

Step-by-step explanation:

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

∴ PQ || AC and PQ = AC (Using mid-point theorem) ... (1)

In ΔADC,

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

∴ RS || AC and RS = AC (Using mid-point theorem) ... (2)

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, it is a parallelogram.

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

In quadrilateral OMQN,

MQ || ON ( PQ || AC)

QN || OM ( QR || BD)

Therefore, OMQN is a parallelogram.

⇒ ∠MQN = ∠NOM

⇒ ∠PQR = ∠NOM

However, ∠NOM = 90° (Diagonals of a rhombus are perpendicular to each other)

∴ ∠PQR = 90°

Clearly, PQRS is a parallelogram having one of its interior angles as 90º.

Hence, PQRS is a rectangle.