Question

Show 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

Proof: ∠ADB = 90° (Angle in a semi-circle is 90°)
∠ADB + ∠ADC = 180° (linear pair)
⇒ ∠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)

Hence, the circle bisects base BC at D.​​​​​​​