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) Consider a $\Delta ABC$. Let two circles be drawn with $AB$ and $AC$ as their respective diameters. Let these circles intersect at point $A$ and another point $D$.
Join $A$ and $D$.
Since $AB$ is a diameter of the first circle,the angle subtended by it at the circumference is $90^{\circ}$.
Therefore,$\angle ADB = 90^{\circ}$ (Angle in a semicircle).
Similarly,since $AC$ is a diameter of the second circle,the angle subtended by it at the circumference is $90^{\circ}$.
Therefore,$\angle ADC = 90^{\circ}$ (Angle in a semicircle).
Adding these two angles,we get:
$\angle ADB + \angle ADC = 90^{\circ} + 90^{\circ} = 180^{\circ}$.
Since the sum of the angles is $180^{\circ}$,the points $B, D,$ and $C$ form a straight line.
Thus,the point of intersection $D$ lies on the third side $BC$.