# Ex.10.6 Q5 Circles Solution - NCERT Maths Class 9

## Question

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

## Text Solution

**What is known?**

Diameter of the circle is the length of one side of rhombus.

**What is unknown?**

Proof of the circle passes through the point of intersection of rhombus diagonals.

**Reasoning:**

The diagonals of rhombus intersect at \(\begin{align}90^\circ.\end{align}\)

**Steps:**

Let \(\begin{align}{ABCD}\end{align}\) be a rhombus in which diagonals are intersecting at point \(\begin{align}{O}\end{align}\) and a circle is drawn while taking side \(\begin{align}{CD}\end{align}\) as its diameter. We know that a diameter subtends \(\begin{align}90^\circ\end{align}\) on the arc.

\(\begin{align}∴{COD }= 90^\circ\end{align}\)

Also, in rhombus, the diagonals intersect each other at \(\begin{align}90^\circ.\end{align}\)

\[\begin {align} \angle {AOB} & =\angle {BOC}\\ &=\angle {COD}\\ & =\angle {DOA} \\ & =90^{\circ}\end {align}\]

Clearly, point \(\begin {align}{O} \end {align}\) has to lie on the circle.