Two circles are drawn on the sides AB and AC of DABC as diameters the circles intersect each other at a point D.

In ΔABC,
AB is the diameter of the circle C(O, r) and AC is the diameter of the circle C'(O', r'). The circles intersect at D.
To Prove:   
∴D lies on BC.
Construction: Join A to D.
Proof: Angle in a semi-circle is 90°.
∴ ∠ADB = 90° and ∠ADC = 90°
Hence, ∠ADB + ∠ADC=180°
⇒ ∠BDC = 180°
Therefore, B, D, C are collinear i.e. D lies on BC.