Heya! In the adjoining figure, P , Q , R and S are the mid - points of the AB , BC , CD and AD respectively. Prove that PQRS is a parallelogram.

Here's a explanation:

Given: ABCD is a rhombus. P,Q,R,S are mid points of AB,BC,CD,DA respectively.

Join: AC and BD

In △ABC

P is mid point of AB and Q is mid point of AC.

Thus, by mid point theorem, PQ∥AC and PQ=

2

1

​

AC

Similarly, In △ACD,

S is mid point of AD and R is mid point of CD.

Thus, by mid point theorem, SR∥AC and SR=

2

1

​

AC

Hence, PQ∥SR and PQ=SR

Similarly, PS=QR and PS∥QR

Thus, the opposite sides of PQRS are equal and parallel.

We know the diagonals of a rhombus bisect each other at right angles.

Now, since AC⊥BD thus, PS⊥PQ (Angle between two lines is same as the angle between their corresponding parallel lines)

Now, the opposite sides of PQRS are equal and parallel and the sides meet each other at right angles. Hence, PQRS is a rectangle.

If PQRS had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if ABCD is a square.

So, the answer is A.

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