Question

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

Answer 1

Join AD.

AB is the diameter.

∴ ∠ADB = 90 (Angle in a semi-circle)

But, ∠ADB + ∠ADC = 180 (linear pair)

∠ADC = 90

In ΔABD and ΔACD,

∠ADB = ∠ADC (each 90)

AB = AC (Given)

AD = AD (Common)

ΔABD ≅ ΔACD (RHS congruence criterion)

BD = DC (C.P.C.T)

Hence, the circle bisects base BC at D.