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.