Question

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.

Answer 1

Consider a ∆ABC. 

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 ∆ABC.