Are you smarter than me?

A litte bit uncomfortably, the statement allows two solutions, the solution the proposer meant to find was that in which the triangle was isosceles with AB=AC. Being that the case, what you have to do is drawing the segment AX in the direction of the ray BA such that AX=AI. This way, BX=BC; the angle AXI is half the angle BAI by a known theorem. Because XBC is isosceles the way we said, and BI is the bisector, we get angle ICB must be equal to angle AXI, that is, half of BAI. But we know CI is bisector of angle ACB. Then you set the equations in terms of BAI, you get the last angle is 45 degrees, or, BAC is 90 degrees.

Extend AI to meet BC at D. Then AD is the median, perpendicular bisector of BC and also an altitude of the triangle.

AD/BD = tan B

ID/BD = tan (B/2)

Subtracting we get: (AD - ID)/BD = AI/BD = tan B - tan (B/2)

Also it is given that AI + AB = BC, so dividing both sides by BC we have:

AI/BC + AB/BC = 1, and since BC = 2 BD

AI/BD + AB/BD = 2

We have derived that AI/BD = tan B - tan (B/2) and from the triangle we can see that AB/BD = sec B

So, tan B - tan (B/2) + sec B = 2

Simplifying this single-variable equation will lead to B = 45 degrees. So, angle BAC = 180 - (B+C) = 180 - 2B = 90

i draw the triangle as described and then i realized it is a right triangle, so i figured out that BC is the hypotenuse and the angle opposite it should be 90 degrees