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.

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