**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.**

**Solution:**

It is given that ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

We need to show that PQRS is a rectangle.

Let's join AC and BD. AC meets SP at H and RQ at F. BD meets PQ at G and SR at E

In , S is the mid-point of DA and R is the mid-point of DC.

Therefore, **by midpoint theorem**, we have and . ** (1)**

Similarly, we have and **by mid-point theorem**. ** (2)**

From **(1)** and **(2)**, we can say that and

is a parallelogram.

**(A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.)**

We have which means that . ** (3)**

Also, we have **by mid-point theorem** which means that . **(4)**

From **(3)** and **(4),** we can say that ERFO is a parallelogram.

**(A quadrilateral is a parallelogram if both pairs of opposite sides are parallel.)**

But, we have because diagonals of rhombus bisect each other at . ** (5)**

And, we have **(Opposite angles of parallelogram are equal) (6)**

From **(5)** and **(6)**, we can say that

It means that PQRS is a parallelogram (**as proved above**) having . It is enough to consider PQRS as rectangle. Therefore, PQRS is a rectangle.