Prove that the angle between the internal bisector of one base angle and the external bisector of the other is - Maths - Lines and Angles

Question

Prove that the angle between the internal bisector of one base
angle and the external bisector of the other is equal to one half
of the vertical angle

Akshita Arora · Accepted Answer

Dear student

Given : In ΔABC BD is the interior angle bisector of ∠B and CD is the exterior angle bisector of ∠C Now In ΔABC ∠A + ∠B + ∠C = 180° ⇒ ∠A = 180° – x – y (1) and In ΔBDE ∠B = $\frac{x}{2}$ $∠ C = y + \frac{180 - y}{2} = \frac{180 + y}{2} and ∠ B + ∠ C + ∠ D = 180 ° \Rightarrow ∠ D = 180 ° - \frac{x}{2} - \frac{180 + y}{2} = \frac{180 - x - y}{2} (2)$ From (1) and (2) $∠ D = \frac{∠ A}{2}$ Regards

Prove that the angle between the internal bisector of one base angle and the external bisector of the other is equal to one half of the vertical angle.