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