If circles are drawn taking two sides of a triangle as diameters,

Solution:



Consider a $\triangle \mathrm{ABC}$.

Two circles are drawn while taking $A B$ and $A C$ as the diameter.

Let they intersect each other at $D$ and let $D$ not lie on $B C$.

Join AD.

$\angle A D B=90^{\circ}$ (Angle subtended by semi-circle)

$\angle \mathrm{ADC}=90^{\circ}$ (Angle subtended by semi-circle)

$\angle B D C=\angle A D B+\angle A D C=90^{\circ}+90^{\circ}=180^{\circ}$

Therefore, BDC is a straight line and hence, our assumption was wrong.

Thus, Point $D$ lies on third side $B C$ of $\triangle A B C$.