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

**Solution:**

Let's draw a figure according to the given question.

Let ABCD be a rhombus in which diagonals intersect at point O, and a circle is drawn by 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 the angle subtended by the diameter on a semicircle is 90°.

Clearly, point O has to lie on the circle.

Thus, the circle passes through the point of intersection of its diagonals O.

**☛ Check: **Class 9 Maths NCERT Solutions Chapter 10

**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.

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