​ 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 = 1/2∠A. ​

Question

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 = \(\frac{1}{2}\)∠A.

Answer 1

Exterior ∠B = (180° - ∠B)

Exterior ∠C = (180° - ∠C)

In ΔABC

∠A + ∠B + ∠C = 180°

\(\frac{1}{2}\)(∠A + ∠B + ∠C) = 180°

\(\frac{1}{2}\)(∠B + ∠C) = 180° - \(\frac{1}{2}\)∠A (i)

In ΔDBC

∠D + ∠DBC + ∠DCB = 180°

∠D + {180° - \(\frac{1}{2}\)(180° - ∠B) - ∠B} + {180° - \(\frac{1}{2}\)(180° - ∠C) - ∠C} = 180°

∠D + 360° – 90° – 90° – (\(\frac{1}{2}\)∠B + \(\frac{1}{2}\)∠C) = 180°

∠D + 180° – 90° - \(\frac{1}{2}\)∠A = 180°

∠D = \(\frac{1}{2}\)∠A

Hence, proved