Question

Prove that the circle drawn on anyone of the equal sides of isosceles triangle as diameter bisects the base

Answer 1

Given: ∆ABC 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


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

∠ADB + ∠ADC = 180°

⇒ ∠ADC = 90°

In ΔABD and ΔACD,

AB = AC (Given)

∠ADB = ∠ADC (90°)

AD = AD (Common)

∴ ΔABD ≅ ΔACD (RHS congruence criterion)

⇒ BD = DC (C.P.C.T)
.... Hope it helps