Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

In △ABC, AB = AC and a circle with AB as diameter is drawn which intersects the side BC at D.

From figure,

∠ADB = 90° [Angle in semi-circle is a right angle.]

Also,

⇒ ∠ADB + ∠ADC = 180° [Linear pair]

⇒ ∠ADC = 180° - ∠ADB

⇒ ∠ADC = 180° - 90° = 90°.

In △ABD and △ACD,

⇒ AB = AC [Given]

⇒ ∠ADB = ∠ADC = 90°

⇒ AD = AD [Common]

∴ △ABD ≅ △ACD by RHS axiom of congruency

∴ BD = CD [By C.P.C.T.]

∴ D is mid-point of BC.

Hence, proved that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.