In the adjoining figure, AD bisects ∠A. Arrange AB, BD and DC in the descending order of their lengths.

∠A = 180° - 60° - 40° = 80°.

Since, AD bisects ∠A,

∠BAD = ∠DAC = $\dfrac{80°}{2}$ = 40°.

∴ ∠ADB = ∠DAC + ∠C (As exterior angle is equal to sum of two opposite interior angles.)

∠ADB = 40° + 40° = 80°.

∴ In △ABD,

∠BAD < ∠ABD < ∠ADB

⇒ BD < AD < AB (side opp. to smaller angle is smaller) …….(i)

Also in △ADC, ∠DAC = 40° = ∠C ⇒ AD = DC …….(ii)

∴ BD < DC < AB or AB > DC > BD.

Hence, sides in descending order are AB > DC > BD.