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.

Construction: Join AD.



Proof:
Since, AC and AB are the diameters of the two circles,
\(\angle{ADB}\) = \(90^\circ\) ...(i)
(Angles in a semi-circle)

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

On adding Equation (i) and (ii), we get,

\(\angle{ADB}\) + \(\angle{ADC}\) = \(90^\circ\) + \(90^\circ\) = \(180^\circ\)

Hence, we can say that, BCD is a straight line.

So, D lies on BC.