<s>[INST] In $\triangle ABC$, if angle bisectors $AE$ and $CD$ meet at incenter $F$, and $|FE|=|FD|$, then the triangle is isosceles or $\angle B=60^\circ$ I was screwing around lately in GeoGebra and I realized something.

Draw a $\triangle ABC$, and let the bisectors for $\angle A$ and $\angle C$ meet sides $BC$ and $AB$ at points $E$ and $D$, respectively. If the angle bisectors meet at the incenter, $F$, and if $FD \cong FE$, then either $\triangle ABC$ must be isosceles or $\angle B$ must be $60^\circ$.

However I was unable to prove why that is. Any help would be appreciated.

 [/INST] Let $AB = c$, $BC = a$, $AC = b$. We will prove that if $EF = FD$, then $ABC$ is either isosceles or $\angle ABC = 60$.
We will first show that $\frac{DF}{FC} = \frac{c}{a+b}$. First, notice that by the angle bisector theorem, we have that $\frac{DF}{FC} = \frac{BD}{BC} = \frac{BD}{a}$. Now, note that $\frac{BD}{c - BD} = \frac{a}{b}$ by the angle bisector theorem, and expanding gives us $BD = \dfrac{\frac{a}{b} \cdot c}{1 + \frac{a}{b}}$. Thus, $\frac{BD}{a} = \dfrac{\frac{c}{b}}{\frac{b+a}{b}} = \frac{c}{b+a}$. Therefore, $\frac{DF}{DC} = \frac{c}{a+b+c}$.
Similarly, we have that $\frac{EF}{FA} = \frac{a}{b+c} \implies \frac{EF}{EA} = \frac{a}{a+b+c}$. Therefore, we have that, since $DF = EF$, $\frac{EA}{DC} = \frac{c}{a} \implies \frac{EA}{AB} = \frac{DC}{BC}$.
Now, note that by the Law of Sines, $\frac{EA}{AB} = \frac{\sin{\angle B}}{\sin \angle{AEB}}$, and $\frac{DC}{BC} = \frac{\sin{\angle{B}}}{\sin{\angle{BDC}}}$, which implies that $\angle AEB = \angle BDC$ or $\angle AEB + \angle BDC = 180$. 
We now consider cases. In the first case, $\angle AEB = 180 - (\angle B + \angle BAE)$ and $\angle BDC = 180 - (\angle B + \angle DCB)$ \implies $\frac{\angle A}{2} = \frac{\angle C}{2}$ which implies $\angle BAC = \angle BCA$, and in the second case we have $360 - 2\angle B - (\angle DCB + \angle BAE) = 180$, and since $\angle DCB + \angle BAE = 90 - \frac{\angle{B}}{2}$ \implies $\frac{3\angle{B}}{2} = 90 \implies \angle{B} = 60$, so we are done.
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