#### Question

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

#### Solution

Join AD.

AB is the diameter.

∴ ∠ADB = 90º (Angle in a semi-circle)

But, ∠ADB + ∠ADC = 180º (linear pair)

⇒ ∠ADC = 90º

In ΔABD and ΔACD,

∠ADB = ∠ADC (each 90º)

AB = AC (Given)

AD = AD (Common)

ΔABD ≅ ΔACD (RHS congruence criterion)

⇒ BD = DC (C.P.C.T)

Hence, the circle bisects base BC at D.

Is there an error in this question or solution?

Solution Show that the Circle Drawn on Any One of the Equal Sides of an Isosceles Triangle as Diameter Bisects the Base. Concept: Tangent to a Circle.