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

(N/A) Let the sides $AB$ and $AC$ of $\Delta ABC$ be the diameters of two circles. These circles intersect each other at points $A$ and $P$.
Draw the common chord $AP$.
Since $AB$ is a diameter,$\angle APB$ is an angle in a semicircle.
$\therefore \angle APB = 90^{\circ}$.
Since $AC$ is a diameter,$\angle APC$ is an angle in a semicircle.
$\therefore \angle APC = 90^{\circ}$.
Adding these two equations,we get:
$\angle APB + \angle APC = 90^{\circ} + 90^{\circ} = 180^{\circ}$.
Since $\angle APB$ and $\angle APC$ are adjacent angles with a common arm $AP$ and their sum is $180^{\circ}$,they form a linear pair.
Therefore,the points $B, P,$ and $C$ must lie on a straight line.
Hence,the point of intersection $P$ of the circles lies on the third side $BC$ of the triangle.