If the 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.

Dear Student,

 

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 ΔABC.

 

 

 

 

Thanks and Regards

Toppers Team