In the adjoining figure, △ABC is equilateral and D is any point on AC. Prove that:

(i) BD > AD

(ii) BD > DC

(i) Since, ABC is an equilateral triangle.

∴ ∠A = ∠B = ∠C = 60°

In △ ABD,

∠ABD = ∠B - ∠DBC

∴ ∠ABD < ∠B

∴ ∠ABD < ∠A (Since, ∠B = ∠A)

∴ AD < BD or BD > AD [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it]

Hence, proved that BD > AD.

(ii) In △ BDC,

∠DBC = ∠B - ∠ABD

∴ ∠DBC < ∠B

∴ ∠DBC < ∠C (∵ ∠B = ∠C)

∴ DC < BD or BD > DC [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it]

Hence, proved that BD > DC.