Question

prove that the angle between the internal bisector of the one base angle and the external bisector of the other equal is equal to one half of the vertical angle angle BEC=1/2angle A

Answer 1

Since angle ACD is an external angle of triangle ABC,

angle ACD = angle B + angle A
=> angle ACD = angle B + angle A
=> angle ACD - angle B = angle A --- (i)

Similarly,

angle ECD = angle EBC + angle BEC
=> angle ECD = angle EBC + angle BEC
=> angle ECD - angle EBC = angle BEC
=> 1/2 angle ACD - 1/2 angle B = angle BEC
=> 1/2 (angle ACD - angle B) = angle BEC
=> 1/2 angle A = angle BEC [from (i)]
=> angle BEC = 1/2 angle A
hence proved

Answer 2

$\textbf{Answer is in Attachment !}$