**Given :** Sides AB and AC of a triangle ABC are diameters of two circles which intersect at D.

**To Prove :** D lies on BC.

**Proof :** Join AD

∠ADB = 90° ...(i) [Angle in a semicircle]

Also, ∠ADC = 90° ..(ii)

Adding (i) and (ii), we get

∠ADB + ∠ADC = 90° + 90°

⇒ ∠ADB + ∠ADC = 180°

⇒ BDC is a straight line.

∴ D lies on BC

Hence, point of intersection of circles lie on the third side BC. **Proved.**