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.

Given: Two circles are drawn with sides AC and AB of $$\triangle{ABC}$$ as diameters. Both circles intersect each other at D.

To prove: D lies on BC.

$$\angle{ADB}$$ = $$90^\circ$$ ...(i)(Angles in a semi-circle)
And $$\angle{ADC}$$ = $$90^\circ$$ ...(ii)(Angles in a semi-circle)
$$\angle{ADB}$$ + $$\angle{ADC}$$ = $$90^\circ$$ + $$90^\circ$$ = $$180^\circ$$