The Total solution for NCERT class 6-12
If circles are drawn taking two sides of a triangle asdiameters, prove that the point of intersection of these circles lie on thethird side.
Consider aΔABC.
Two circles are drawn while taking AB and AC as thediameter.
Let they intersect each other at D and let D not lie onBC.
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, ourassumption was wrong.
Thus, Point D lies on third side BC of ΔABC.