In the given figure, AD bisects ∠A. If ∠B = 60°, ∠C = 40°, then arrange AB, BD and DC in ascending order of their lengths.

Given,

AD bisects ∠A.

⇒ ∠BAD = ∠CAD = x (let)

In △ABC,

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 60° + 40° = 180°

⇒ ∠A + 100° = 180°

⇒ ∠A = 180° - 100°

⇒ ∠A = 80°

⇒ ∠BAD + ∠CAD = 80°

⇒ x + x = 80°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$

⇒ x = 40°

⇒ ∠BAD = ∠CAD = 40°

In △ABD,

By angle sum property of triangle,

⇒ ∠BAD + ∠B + ∠ADB = 180°

⇒ 40° + 60° + ∠ADB = 180°

⇒ 100° + ∠ADB = 180°

⇒ ∠ADB = 180° - 100°

⇒ ∠ADB = 80°

We know that,

In a triangle larger angle has larger side opposite to it.

Since,

∠ADB > ∠ABD > ∠BAD

∴ AB > AD > BD …….(1)

From figure,

⇒ ∠ADB + ∠ADC = 180° (Linear pair)

⇒ 80° + ∠ADC = 180°

⇒ ∠ADC = 180° - 80°

⇒ ∠ADC = 100°

Since,

⇒ ∠DAC = ∠ACD (Both equal to 40°)

∴ AD = DC (Sides opposite to equal angles are equal)

Substituting value of AD in equation (1), we get :

⇒ AB > DC > BD

⇒ BD < DC < AB.

Hence, BD < DC < AB.