# 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.

**☛ Related Questions:**

- ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
- AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
- Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° - 1/2 A, 90° - 1/2 B , 90° - 1/2 C.
- Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

visual curriculum