ABC is a triangle in which AB=AC and I is the incentre of triangle

given BC=AB+AI find angle BAC

ΔABC is isosceles as AB=AC, 

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
B = 45  = C
A = 90