. 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.

Construction: Join AD.
Proof: AB is diameter of circle and ∠ADB is Formed in semi-circle.
Hence, ∠ADB = 90° …(1) [∵ Angle in a semicircle is a right angle]
Similarly, AC is diametar and ∠ADC is formed in semi- circle.
Hence, ∠ADC = 90° …(2) [∵ Angle in a semicircle is a right angle]
Here, ∠ADB + ∠ADC = 90° + 90° = 180°
∠ADB and ∠ADC are forming linear pair. Therefore BDC is a straight line.
Hence, the point D lies on third side BC.