# Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

**Solution:**

The diagonals of the rhombus intersect at 90º.

Let ABCD be a rhombus in which diagonals intersect at point O, and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.

Therefore, ∠COD = 90°

Also, in the rhombus, the diagonals intersect each other at 90°.

∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

But ∠COD is 90° and this can only happen on a semicircle with diameter DC since angle subtended by the diameter in a semicircle is 90°.

Clearly, point O has to lie on the circle.

**Video Solution:**

## Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

### Maths NCERT Solutions Class 9 - Chapter 10 Exercise 10.6 Question 5:

**Summary:**

We have proved that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.