Consider a ΔABC.
Two circles are drawn while taking AB and AC as the diameter.
Let they intersect each other at D and let D not lie on BC.
Join AD.
∠ADB = 90° (Angle subtended by semi-circle) ∠ADC
= 90° (Angle subtended by semi-circle) ∠BDC =
∠ADB + ∠ADC = 90° + 90° = 180°
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, Point D lies on third side BC of ΔABC.