Extend $AC$ past $A$ to $P$ so that $AP = AI$. Then $PC = PA + AC = AI + AC = BC$, so triangle $PBC$ is isosceles. Since $\angle ACI = \angle BCI$, triangles $BIC$ and $PIC$ are congruent, so $\angle IPC = \angle IBC = \frac{B}{2}$. Then $\angle AIP = \angle API = \frac{B}{2}$. Also, $\angle BAI = \frac{A}{2}$ and $\angle PAB = 180^\circ - A$, so adding up all the angles in triangle $PAI$, we get \[\frac{B}{2} + 180^\circ - A + \frac{A}{2} + \frac{B}{2} = 180^\circ.\]This simplifies to $A = 2B$. [asy][asy] unitsize(1 cm); pair A, B, C, I, P; B = (0,0); C = (5,0); A = extension(B, B + dir(50), C, C + dir(180 - 30)); I = incenter(A,B,C); P = C + 5*dir(180 - 30); draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(A--P--B); draw(B--I--C); draw(A--I--P); label("$A$", A, NE); label("$B$", B, SW); label("$C$", C, SE); dot("$I$", I, S); label("$P$", P, N); [/asy][/asy]

Let $K$ be a point on $BC$ such that $KC=AC$. Note that $$AI=BC-AC=BC-KC=BK.$$Moreover, $$\angle AIB=90^{\circ}+\frac{1}{2}\angle ACB=180^{\circ}-\angle AKC=\angle AKB.$$Thus, $AIKB$ is an isosceles trapezoid, so $$\angle IAB=\angle KBA\implies \frac{1}{2}\angle BAC=\angle ABC\implies BAC=2\angle ABC.$$