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.

 Solution:

First draw a triangle ABC and then two circles having diameter as AB and AC respectively.

We will have to now prove that D lies on BC and BDC is a straight line.

Proof:

We know that angle in the semi-circle are equal

So, ∠ ADB = ∠ ADC = 90°

Hence, ∠ ADB+∠ ADC = 180°

∴ ∠ BDC is straight line.

So, it can be said that D lies on the line BC.