Question

prove that the circle drawn on any one equal side of the isosceles triangle as diameter bisect 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)