# 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 = (180o - B)

Exterior C = (180o - C)

In

A + B + C = 180o

(A + B + C) = 180o

(B + C) = 180o - A (i)

In

D + DBC + DCB = 180o

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

D + 360o – 90o – 90o – (B + C) = 180o

D + 180o – 90o - A = 180o

D = A

Hence, proved

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