Question

If circles are drawn taking two sides of a triangle as diameters, prove that the point intersection of these circles lie on the third side.

Answer 1

Consider a $\Delta ABC$
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.
$\angle ADB={90}^{\circ }$ (Angle subtended by semi - circle)
$\angle ADC={90}^{\circ }$ (Angle subtended by semi - circle)
$\angle BDC=\angle ADB+\angle ADC={90}^{\circ }+{90}^{\circ }={180}^{\circ }$
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, point D lies on third side BC of $\Delta ABC$ .