Question

ABCD is a rhombus P,Q,R and S mid points of AB,BC,CD and AD .Show that quadrilateral PQRS is a rectangle.

Answer 1

Hello Mate!

Given : ABCD is rhombus where PQRS are formed by joining mid points.

To prove : PQRS is a rectangle.

To construct : Join AC and BD.

Proof : Since RQ are formed by joining mid points so, QR || BD and QR = ½ BD _(i)

Since PS are formed by joining mid points so, PS || BD and PS = ½ BD _(ii)

From (i) and (ii) we get,

QR || PS and QR = PS

Hence a pair if side is equal and parallel so PQRS is ||gm.

Now, since QR || BD so MR || ON __(iii)

Similarly, NR || AC __(iv) [ Because SR || AC, since SR is formed by joining mid points ]

From (iii) and (iv) we get, MONR is a ||gm.

As we know that diagonals of rhombus bisect at 90° so < MON = 90°

Now, since MONR is ||gm so its opposite angles will be equal. So,

< MRN = 90°

Or < QRS = 90°

Hence PQRS is such ||gm whose one if the angle ( < QRS ) is 90° so PQRS is a rectangle.

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ.

Have great future ahead!

Answer 2

here is your answer OK ☺☺☺☺☺☺☺☺



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.