ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = ∠A.

Exterior ∠B = (180^o - ∠B)

Exterior ∠C = (180^o - ∠C)

In

∠A + ∠B + ∠C = 180^o

(∠A + ∠B + ∠C) = 180^o

(∠B + ∠C) = 180^o - ∠A (i)

In

∠D + ∠DBC + ∠DCB = 180^o

∠D + {180^o - (180^o - ∠B) - ∠B} + {180^o - (180^o - ∠C) - ∠C} = 180^o

∠D + 360^o – 90^o – 90^o – (∠B + ∠C) = 180^o

∠D + 180^o – 90^o - ∠A = 180^o

∠D = ∠A

Hence, proved