Question 252016
The angle BAC that satisfies the relation BC=AB+AI is the right angle.
To get to this solution with some trigonometry, let's call x the angle BAC and l the lenght of AB..
As the triangle ABC is isosceles and the sum of the internal angles of any triangle is 2 right angles, then the angle {{{ABC=ACB=(PI-x)/2=PI/2 -X/2}}}.
Let me remind that the Incenter of a triangle can be found as the intersection of any two internal angle bisectors.
Let's call H the intersection of the bisector of BAC and BC.
AH is also the altitude relative to the base BC. So ABH is a right triangle (right in H).
For the definition of sine {{{BH=BA*sin(x/2)=l *sin(x/2)}}}
For the  Pythagorean theorem {{{AH=sqrt(l^2-l^2*sin^2(x/2))=l*sqrt(1-sin^2(x/2))=l*sqrt(cos^2(x/2))=l*cos(x/2)}}}
The angle IBH is half of ABH that is {{{IBH=ABC / 2= PI/4-x/4}}}
For the definition of tangent {{{IH=BH*tan(IBH)=l*sin(x/2)*tan(PI/4-x/4)}}}
Finally {{{AI=AH-IH=l*cos(x/2)-l*sin(x/2)*tan(PI/4-x/4)}}}
Therefore, as {{{BC=2*BH}}}, the relation {{{BC=AB+AI}}} can be written {{{2*l *sin(x/2)=l+l*cos(x/2)-l*sin(x/2)*tan(PI/4-x/4)}}} and with some trivial passage
{{{2*sin(x/2)=(1+cos(x/2)-sin(x/2)*tan(PI/4-x/4))}}}
Still Algebra:
{{{sin(x/2)*(2+tan(PI/4-x/4))=1+cos(x/2)}}}
Now we apply the sum identity of the tangent {{{tan(alpha-beta)=(tan(alpha)-tan(beta))/(1+tan(alpha)*tan(beta))}}} to obtain:
{{{sin(x/2)*(2+(tan(PI/4)-tan(x/4))/(1+tan(PI/4)*tan(x/4)))=1+cos(x/2)}}}
Known that {{{tan(PI/4)=1}}}, we obtain:
{{{sin(x/2)*(2+(1-tan(x/4))/(1+tan(x/4)))=1+cos(x/2)}}}
Now let's apply the half angle formula for the tangent {{{tan(alpha/2)=(1-cos(alpha))/sin(alpha)}}}
{{{sin(x/2)*(2+(1-(1-cos(x/2))/sin(x/2))/(1+(1-cos(x/2))/sin(x/2)))=1+cos(x/2)}}}
Again Algebra and provided that {{{sin(x/2)<>0}}} i.e {{{x<>0}}} we obtain:
{{{sin(x/2)*(2+(sin(x/2)-1+cos(x/2))/(sin(x/2)+1-cos(x/2)))=1+cos(x/2)}}}
then
{{{sin(x/2)*(2-(1-sin(x/2)-cos(x/2))/(1+sin(x/2)-cos(x/2)))=1+cos(x/2)}}}
and
{{{sin(x/2)*(2+2*sin(x/2)-2*cos(x/2)-1+sin(x/2)+cos(x/2))/(1+sin(x/2)-cos(x/2)))=1+cos(x/2)}}}
and provided that {{{(1+sin(x/2)-cos(x/2))<>0}}}
{{{sin(x/2)*(1+3*sin(x/2)-cos(x/2))=(1+cos(x/2))*(1+sin(x/2)-cos(x/2))}}}
and
{{{sin(x/2)+3*sin^2(x/2)-sin(x/2)*cos(x/2))=1+cos(x/2)+sin(x/2)+sin(x/2)*cos(x/2)-cos(x/2)-cos^2(x/2)}}}
and 
{{{3*sin^2(x/2)-sin(x/2)*cos(x/2))=1-cos^2(x/2)+sin(x/2)*cos(x/2)}}}
and 
{{{3*sin^2(x/2)-sin(x/2)*cos(x/2))=sin^2(x/2)+sin(x/2)*cos(x/2)}}}
and
{{{2*sin^2(x/2)-2*sin(x/2)*cos(x/2))=0}}}
and
{{{2*sin(x/2)*(sin(x/2)-cos(x/2))=0}}}
The first term cannot be zero for the previous condition, so the possible solution must satisfy:
{{{sin(x/2)-cos(x/2)=0}}}
that happens when {{{x/2=PI/4}}} i.e. {{{x=PI/2}}}.
The general solution would be {{{x=2*PI*n-3/2*PI}}} with n any integer, but for the geometric nature of the problem we can rest on {{{n=1}}}.

Eventually we check the solution that is that in case of a right/isosceles triangle the relation BC=AB+AI is satisfied.
With the same notation above:
{{{BC=sqrt(2*l^2)=l*sqrt(2)=l*(1+(sqrt(2)-1))}}}
BH is such that {{{l^2=2*BH^2}}} therefore {{{BH=l/sqrt(2)}}}
As {{{ABC=PI/2}}} then {{{ABH=PI/4}}} and {{{IBH=PI/8}}}, then {{{IH=BH*tan(PI/8)=BH*(1-cos(PI/4))/sin(Pi/4)=BH*(1-sqrt(2)/2)/(sqrt(2)/2)=BH*(2-sqrt(2))/sqrt(2)}}}
But {{{AH=BH=l/sqrt(2)}}} and {{{AI=AH-IH=l/sqrt(2)-l/sqrt(2)*(2-sqrt(2))/sqrt(2)=l*(sqrt(2)-1)}}}
So {{{AB+AI=l+l*(sqrt(2)-1)=l*sqrt(2)}}} and {{{BC=2*BH=2*l/sqrt(2)=l*sqrt(2)}}}
Quod Demonstrandum Erat