Question

Prove that the circle drawn with any equal side 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

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)

:) Hope this helps!!!