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.