Question

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

• A. True
• B. False

Answer 1

The correct option is A

True

Given- $\Delta$ ABC in which AB = AC and AB as diameter, a circle is drawn which intersects BC at D.

To Prove-BD = DC

Construction- Join AD

Proof- AB is the diameter

$\therefore \angle ADB={90}^0$ (Angle in a semi-circle)

But $\angle ADB\angle ADC={180}^0$ (A linear pair)

$\angle ADC={90}^0$

Now in right angled $\Delta$s ABD and ACD,

Hypotenuse AB = AC(given) Side AD = AD (Common)

$\therefore \Delta ABD\cong \Delta ACD$

BD = DC (RHS postulate) BD =DC (C.P.C.T.)

Hence the circle bisects base BC at D. Q.E.D.