If circles are drawn taking two sides of a triangle as diameters,check that the point of intersection of these circles lie on the third side.
Proof:
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 \(AD\).
\(∠ADB =\) \(^°\) (Angle subtended by semi-circle)
\(∠ADC =\) \(^°\) (Angle subtended by semi-circle)
\(∠BDC = ∠ADB + ∠ADC =\) \(^°\)
Therefore, \(BDC\) is a straight line and hence, our assumption was wrong.
Thus, Point \(D\) lies on third side \(BC\) of \(ΔABC\).
Hence, proved.