Question

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

Answer 1

Given: ΔABC is an isosceles triangle with AB = AC. A circle is drawn taking AB as the diameter which intersects the side BC at D.

To prove: BD = DC

Construction: Join AD

http://www.meritnation.com/img/userimages/mn_images/image/zzzz.png

Proof: ∠ADB = 90° (Angle in a semi-circle is 90°)

∠ADB + ∠ADC = 180°

⇒ ∠ADC = 90°

In ΔABD and ΔACD,

AB = AC   (Given)

∠ ADB = ∠ ADC (Proved)

AD = AD (Common)

⇒ ΔABD  ΔACD   (RHS congruence criterion)

⇒BD = DC (C.P.C.T)