$\textbf{Lemma:}$ In a triangle $\triangle{ABC}$, let the incircle and the angle bisector from $A$ intersect $BC$ at $X,D$, respectively. Then $BX\cdot XC\leq BD\cdot DC$, equality case being $AB=AC$. $\textbf{Proof:}$ WLOG suppose that $AB>AC$ (the case $AB=AC$ is trivial), $\angle{ADC}=\angle{DAB}+\angle{ABC}=\frac{\angle{A}}{2}+\angle{B}<90$, so $X$ lies between $C$ and $D$ (Note that the incenter $I$ lies on $[AD]$ and $IX\perp BC$). Let $M$ be the midpoint of $[BC]$, $M$ lies between $D$ and $B$, thus $MD<MX$. $BX\cdot XC=(BM+MX)\cdot (CM-MX)=BM^2-MX^2$, similarly $BD\cdot DC=BM^2-MD^2$, the result follows. Let $I$ be the incenter of $\triangle{ABC}$ It is known that $IB\perp BI_C$, thus $DF^2=BD^2+BF^2$ From $\triangle{CBF}$ with $\angle{CBF}=90+\frac{\angle{B}}{2}$ we get $CF^2=CB^2+BF^2-2\cdot CB\cdot BF\cdot \cos (90+\frac{\angle{B}}{2})$ Observe that $2\cdot BF=BI_C$ and $-BI_C\cdot\cos (90+\frac{\angle{B}}{2})=BI_C\cdot\cos (90-\frac{\angle{B}}{2})=\frac{AB+AC-BC}{2}$ (Let the excircle touch $BC$ at $T$, then it is known that $BT=\frac{AB+AC-BC}{2}$, and one can see that $\frac{BT}{BI_C}=\cos\angle{ABI_C}=\cos (90-\frac{\angle{B}}{2})$). It follows that $CF^2=BC^2+BF^2+BC\cdot\frac{AB+AC-BC}{2}=BD^2+BF^2+CE^2$, thus $BD^2+CE^2=\frac{BC^2+BC\cdot AB+BC\cdot AC}{2} (1)$ It is known that $BD^2=AB\cdot BC-AD\cdot DC$ and $CE^2=AC\cdot CB-AE\cdot EB$, inserting in $(1)$ gives $AD\cdot DC+AE\cdot EB=BC\cdot\frac{AB+AC-BC}{2}$ Let the incircle touch $AC,AB$ in $X,Y$, respectively. It is known that $AX=AY=\frac{AB+AC-BC}{2}$, $CX=\frac{AC+BC-AB}{2}$ and $BY=\frac{AB+BC-AC}{2}$, it follows that $AX\cdot XC+AY\cdot YB=BC\cdot \frac{AB+AC-BC}{2}=AD\cdot DC+AE\cdot EB$ On the other hand, using the lemma. we get $AD\cdot DC\geq AX\cdot XC$ and $AE\cdot EB\geq AY\cdot YB$, thus the equality conditions $AB=BC$ and $AC=BC$ must be fulfilled, the result follows.

Let $K$ and $L$ be the feet of the perpendicular from $F$ and $I_c$ to $BC$. $DF^2=FB^2+BD^2$ so we know that $CF^2=CE^2+FB^2+BD^2\Leftrightarrow CF^2-FB^2=CE^2+BD^2$. $I_cF=FB$ so $LK=KB$. $\angle FKB=90^{\circ}\Rightarrow CE^2+BD^2=CF^2-FB^2=CK^2-BK^2=(CK-BK)(CK+BK)=(CK-BK)(CK+KL)=CB.CL$ Let $BC=4a,CA=4b$ and $AB=4c$. Then $CB.CL=4a(2a+2b+2c)=8(a^2+ab+ac)$ and $CE^2+BD^2=(16ab-\frac{16b^2ac}{(a+c)^2})+(16ac-\frac{16c^2ab}{(a+b)^2})$. $CE^2+BD^2=CB.CL\Leftrightarrow (16ab-\frac{16b^2ac}{(a+c)^2})+(16ac-\frac{16c^2ab}{(a+b)^2})=8(a^2+ab+ac)\Leftrightarrow (2ab-\frac{2b^2ac}{(a+c)^2})+(2ac-\frac{2c^2ab}{(a+b)^2})=a^2+ab+ac\Leftrightarrow ab+ac-a^2=\frac{2b^2ac}{(a+c)^2}+\frac{2c^2ab}{(a+b)^2}\Leftrightarrow b+c-a=\frac{2b^2c}{(a+c)^2}+\frac{2c^2b}{(a+b)^2}$. Let $a=x+y,b=y+z$ and $c=z+x$. Then the equality becomes $2z=\frac{2(y+z)^2(z+x)}{(2x+y+z)^2}+\frac{2(z+x)^2(y+z)}{(2y+x+z)^2}\Rightarrow z=\frac{(y+z)^2(z+x)}{(2x+y+z)^2}+\frac{(z+x)^2(y+z)}{(2y+x+z)^2}$. Now, let's show that $\frac{(y+z)^2(z+x)}{(2x+y+z)^2}\ge \frac{zy}{x+y}$. $\frac{(y+z)^2(z+x)}{(2x+y+z)^2}\ge \frac{zy}{x+y}\Leftrightarrow \frac{(x+y)(z+x)}{(2x+y+z)^2}\ge \frac{zy}{(y+z)^2}\Leftrightarrow \frac{(2x+y+z)^2}{(x+y)(z+x)}\leq \frac{(y+z)^2}{yz}\Leftrightarrow \frac{(2x+y+z)^2-4(x+y)(z+x)}{(x+y)(z+x)}\leq \frac{(y+z)^2-4yz}{yz}\Leftrightarrow \frac{(y-z)^2}{(x+y)(z+x)}\leq \frac{(y-z)^2}{yz}\Leftrightarrow (y-z)^2(\frac{x^2+xy+xz}{(x+y)(z+x)yz})\ge 0$, which is correct. Similarly, $\frac{(y+z)(z+x)^2}{(2y+x+z)^2}\ge \frac{zx}{x+y}$. In conclusion, $z=\frac{(y+z)^2(z+x)}{(2x+y+z)^2}+\frac{(y+z)(z+x)^2}{(2y+x+z)^2}\ge \frac{zy}{x+y}+\frac{zx}{x+y}=z$. Thus, we must have equality for all inequalities. $(y-z)^2(\frac{x^2+xy+xz}{(x+y)(z+x)yz})=0\Rightarrow (y-z)^2=0\Rightarrow y=z$. Similarly, $y=x$. Then $a=b=c$. Hence, $ABC$ is an equilateral triangle.