Question

Question 10
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 $\Delta ABC$.
Two circles are drawn while taking AB and AC as the diameter.
Let they intersect each other at D.
Assume that the point of intersection 'D' doesn't lie on BC and join AD.
$\angle ADB={90}^{\circ }$ (Angle subtended by semi-circle)
$\angle ADC={90}^{\circ }$ (Angle subtended by semi-circle)
$\angle BDC=\angle ADB+\angle ADC={90}^{\circ }+{90}^{\circ }=180$
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, Point D lies on third side BC of $\Delta ABC$.