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.

Two circles are drawn taking sides AB and AC as diameter.

⇒ ∠ADB = ∠ADC = 90° (Angles in the semi circle is a right angle.)

⇒ ∠ADB + ∠ADC = 90° + 90°

⇒ ∠ADB + ∠ADC = 180°

∴ BDC is straight line.

Thus, D lies on BC.

Hence, proved that the point of intersection of circles lie on the third side.