Question

In Triangle ABC, AB = BC and BC = AB + AI, where I is the incentre of Triangle ABC. Then, the measure of Angle A is ?
(Figure given in the picture attached)
Thanks in advance! Please help me out!

Answer 1

This is just relation between in-radius and circum-radius in terms of trigonometric identites.

Answer 2

ΔABC is isosceles as AB=AC, $\ \angle{B}=\angle{C},\ \angle{B}=90^0-\angle{A}/2,\ \ Sin(A/2)=Cos(B)$

In ΔABD:  cosB = a/2b      ----- (1)
In ΔAIE:    r = AI sin A/2 = AI cosB = AI *a/2b   using (1)   ...... (2)
In ΔBID:    r = a/2 * tan(B/2)
       =>  AI = 2br /a = b tan(B/2)              using (2)     ...... (3)

given   a = b + AI  = b [1+tan(B/2) ]            using (3)

from (1) & (2):          1+ tan(B/2)] = 2*CosB = [1 - tan²(B/2)]/[1+tan²(B/2)]

=>   tan^2 B/2 + 2 tan B/2 - 1 = 0
=>   tan B/2 = √2 - 1=>   cosB = 1/√2
=>   $\angle{B}= 45^0 = \angle{C},\ \ \ \angle{A}=90^0$