For I the incenter in △ABC, if AB+IC=AC+IB, then △ABC is isosceles. 
Let $I$ be the center of the inscribed circle in $\triangle ABC$. Prove that if $$AB+IC=AC+IB$$ then the triangle is isosceles!


Another way.
In the standard notation we have
$$c+\frac{r}{\sin\frac{\gamma}{2}}=b+\frac{r}{\sin\frac{\beta}{2}}$$ or
$$c+\frac{\frac{2S}{a+b+c}}{\sqrt{\frac{1-\frac{a^2+b^2-c^2}{2ab}}{2}}}=b+\frac{\frac{2S}{a+b+c}}{\sqrt{\frac{1-\frac{a^2+c^2-b^2}{2ac}}{2}}}$$ or
$$b-c=\frac{4S\sqrt{ab}}{(a+b+c)\sqrt{(a+c-b)(b+c-a)}}-\frac{4S\sqrt{ac}}{(a+b+c)\sqrt{(a+b-c)(b+c-a)}}$$ or
$$b-c=\frac{4S\sqrt{a}}{(a+b+c)\sqrt{b+c-a}}\left(\frac{\sqrt{b}}{\sqrt{a+c-b}}-\frac{\sqrt{c}}{\sqrt{a+b-c}}\right)$$ or
$$b-c=\frac{\sqrt{a}\left(\sqrt{b(a+b-c)}-\sqrt{c(a+c-b)}\right)}{\sqrt{a+b+c}}$$ or
$$b-c=\frac{\sqrt{a}\left(b(a+b-c)-c(a+c-b)\right)}{\sqrt{a+b+c}\left(\sqrt{b(a+b-c)}+\sqrt{c(a+c-b)}\right)}$$ or
$$b-c=\frac{\sqrt{a}(b-c)(a+b+c)}{\sqrt{a+b+c}\left(\sqrt{b(a+b-c)}+\sqrt{c(a+c-b)}\right)},$$ which gives $b=c$ or
$$\sqrt{b(a+b-c)}+\sqrt{c(a+c-b)}=\sqrt{a(a+b+c)}$$ or
$$2\sqrt{bc}=\sqrt{(a+b-c)(a+c-b)}$$ or
$$(b+c)^2=a^2,$$ which is impossible.
Id est, $b=c$ and we are done! 
