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.
Let $a$ and $c$ be the half angles,

$$ a = \frac{∠A}{2}, \space  c = \frac{∠C}{2}.$$

Apply the sine rule to the triangles ΔBDF and ΔBEF, respectively,

$$ DF = BF \frac{\sin(∠B/2)}{\sin(2a+c)}\tag{1}$$
$$ EF = BF \frac{\sin(∠B/2)}{\sin(2c+a)}\tag{2}$$

where ∠BDC = $2a + c$ and ∠BEA = $2c + a$ are recognized. 

Take the ratio of (1) and (2),

$$ \frac{DF}{EF} = \frac{\sin(2c+a)}{\sin(2a+c)}$$

If DF = EF, the equation yields,

$$\sin(2c+a)=\sin(2a+c).$$

which gives two solutions,

$$2c+a = 2a+c\tag{3}$$

$$2c+a = 180^\circ - (2a+c)\tag{4}$$


The solution (3) leads to $a = c$, which is the isosceles triangle case.
The solution (4) leads to $a + c = 60^\circ$ and in turn ∠B = $180^\circ -2(a+c)=60^\circ $, which is the 60-degree case.


Hence, both cases are proved.
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?