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!

 A: Extend $AB$ across $B$ for lenght $CI$ and we get a point $D$ 
and 
extend $AC$ across $C$ for lenght $BI$ and we get a point $E$. 
So $AD = AE$ and $AI $ bisect angle at $A$, so triangles $ADI$ and $AEI$ are congruent by (sas). 
But then $DI = EI$ so triangles $BDI$ and $DEI$ are also congurent by (sss).
But since $\angle IDB = \angle IEC$ these two triangles are also isosceles, so $BI=CI$ and thus $AB= AC$.
A: 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! 
A: Let us assume that $ABC$ is the orthic triangle of $DEF$: we have $BC= R\sin(2D)$ and the incenter of $ABC$ is the orthocenter of $DEF$, hence $IA=2R\cos(E)\cos(F)$. The statement can be written in the form
$$\sin(F)\cos(F)+\cos(D)\cos(E)= \sin(E)\cos(E)+\cos(D)\cos(F)$$
or in the equivalent form
$$8 \sin\left(\frac{E}{2}-\frac{F}{2}\right)\sin\left(\frac{E}{2}-\frac{\pi }{4}\right)\sin\left(\frac{F}{2}-\frac{\pi }{4}\right)\sin\left(\frac{E}{2}+\frac{F}{2}-\frac{\pi }{4}\right)\sin\left(\frac{E}{2}+\frac{F}{2}+\frac{\pi }{4}\right)=0.$$
We get $E=F$, hence $B=C$.
