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

Let circle be drawn on one of the equal sides AB of the isosceles triangle ABC as shown in the figure below:

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

But,

⇒ ∠ADB + ∠ADC = 180° [Linear pair]

⇒ 90° + ∠ADC = 180°

⇒ ∠ADC = 180° - 90°

⇒ ∠ADC = 90°.

In ∆ABD and ∆ACD, we have

⇒ ∠ADB = ∠ADC [Each 90°]

⇒ AB = AC [Given]

⇒ AD = AD [Common]

Hence, ∆ABD ≅ ∆ACD by RHS congruence criterion.

By, C.P.C.T we get :

BD = DC

Hence, the circle bisects base BC at D.