<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] In the standard notation by the angle bisector theorem we obtain:
$$\frac{FE}{AF}=\frac{CF}{AC}=\frac{\frac{ab}{b+c}}{b}=\frac{a}{b+c}.$$
Thus, $$\frac{FE}{AE-FE}=\frac{a}{b+c},$$ which gives
$$FE=\frac{a}{a+b+c}\cdot AE=\frac{a}{a+b+c}\cdot\frac{2bc\cos\frac{\alpha}{2}}{b+c}=\frac{2abc\cos\frac{\alpha}{2}}{(a+b+c)(b+c)}.$$
Similarly, $$FD=\frac{2abc\cos\frac{\gamma}{2}}{(a+b+c)(a+b)}.$$ 
Id est, 
$$\frac{2abc\cos\frac{\alpha}{2}}{(a+b+c)(b+c)}=\frac{2abc\cos\frac{\gamma}{2}}{(a+b+c)(a+b)}$$ or
$$a(b+c-a)(a+b)^2=c(a+b-c)(b+c)^2$$ or
$$(a-c)(a+b+c)(a^2+c^2-ac-b^2)=0.$$
Can you end it now?
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