Question

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

Answer 1

Given : In ΔABC BD is the interior angle bisector of ∠B and CD is the exterior angle bisector of ∠C. *Proof* : Now In ΔABC ∠A + ∠B + ∠C = 180° ⇒ ∠A = 180° – x – y  ....  (1); and In ΔBDE-- ∠B = ∠C= y+ (180-y)/2 = (180+y)/2. And, ∠B+ ∠C+ ∠D=180°=> ∠D= 180- x/2 - (180-y)/2 => (180-x-y)/2 ...(2). So, from (1) & (2)---- ∠D= ∠A/2.

Answer 2

$\textbf{Answer is in Attachment !}$