# Ex.10.5 Q10 Circles Solution - NCERT Maths Class 9

## Question

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

## Text Solution

**What is given ?**

Two circles are drawn taking two sides of a triangle as diameters.

**What is unknown?**

To prove that point of intersection of the **\(2\)** circles lie on the third side.

**Reasoning:**

Angle in a semicircle is a right angle. By using this fact we can show that \(BDC\) is a line which will lead to the proof that point of intersection lie on the third side.

**Steps:**

Since angle in a semicircle is a right angle, we get:

\( \begin{align} \angle {ADB}&=90^{\circ} \text { and } \angle {ADC}=90^{\circ} \end{align}\)

\(\begin{align} \angle {ADB}+\angle {ADC}&=90^{\circ}+90^{\circ} \\ \Rightarrow \;\; \angle {ADB}+\angle {ADC}&=180^{\circ} \end{align}\)

\(\Rightarrow \quad{BDC} \text { is a straight line. }\)

∴ \({D}\) lies on \({BC}\)

Hence, point of intersection of circles lie on the third side \(BC.\)