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}$ and $ c = \frac{C}{2}$ respectively. Apply the sine rule to the triangles ΔBDF and ΔBEF
$$ DF = BF \frac{\sin\frac B2}{\sin(2a+c)}\tag{1}$$ $$ EF = BF \frac{\sin\frac B2}{\sin(2c+a)}\tag{2}$$
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.
I am trying to give a simple, purely geometric proof. Here is first a picture, notations are slightly changed, so that it is simpler to type formulas blindly:
$\Delta ABC$ is a triangle, $AA'$, $BB'$, $CC'$ are its angle bisector cevians, $I$ is their intersection, the incenter. We assume $IA'=IC'$, and that the perpendiculars $IU$ and $IS$ from $I$ on $BA$ and respectively $BC$, $U\in BA$, $S\in BC$, are so that one is inside the angle $\widehat{A'IC'}$, one outside. In our case, without loss of generality, $S$ is on the segment $BA'$, and $U$ is not on the segment $BC'$.
We assume $$IA'=IC'\ .$$
Then $\color{red}{\hat B=60^\circ}$.
Proof: The triangles $\Delta ISA'$ and $\Delta IUC'$ are congruent, having right angles in $S$, respectively $U$, and $IS=IU$, and the given relation $IA'=IC'$. So their two angles in $I$ are equal. So the two angles in $I$ obtained by adding $\widehat {SIC'}$ are equal. This implies: $$ \pi-\frac 12 \hat A-\frac 12C = \color{blue}{ \widehat{AIC} = \widehat{A'IC'} = \widehat{SIU} } = \pi-\hat B\ . $$ The blue part is the idea, the "mule" to connect the given information with the needed angle information. We obtain immediately $\hat B =\frac 12(\hat A+\hat C)=\frac12(\pi-B)$, so $3\hat B=\pi$.
$\square$
The other case, when the perpendiculars from $I$ on the sides $BA$, $BC$ are one "the same part" is arguably simpler. After drawing the picture and using the same notations, after using again the congruence of the two triangles $\Delta IA'S$ and $\Delta IC'U$, we get this time $$ \frac 12(\hat B+\hat A) = \widehat{ABI} + \widehat{IAB} = \color{blue} { \widehat{BIA'} = \widehat{BIC'} } = \widehat{IBC} + \widehat{ICB} =\frac 12(\hat B+\hat C) \ . $$ So $\hat A=\hat C$.
$\square$
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.
Given a triangle as described in the question, where $FD \cong FE.$
We first show that $\angle FDB$ and $\angle FEB$ are either congruent or supplementary angles. There are two cases:
Case 1: The circle is tangent to $AB$ at $D.$ Then $FD$ is a radius of the circle, and $FD \cong FE$ implies that $FE$ also is a radius of the circle, hence the circle is tangent to $BC$ at $E.$ Then $\angle FDB$ and $\angle FEB$ both are right angles, hence are both congruent and supplementary in Case 1.
Case 2: The circle is not tangent to $AB$ at $D.$ Let $H$ be the point of tangency with $AB$ and let $K$ be the point of tangency with $BC.$ Then $\lvert FD\rvert > \lvert FH\rvert.$ Further, $FD \cong FE$ and $FH \cong FK$ imply that $\lvert FE\rvert > \lvert FK\rvert.$
Then $\triangle FHD$ and $\triangle FEK$ are right triangles with a congruent leg and congruent hypotenuse, so $\triangle FHD \cong \triangle FKE$ with $\angle FDH \cong \angle FEK.$
But $\angle FDB$ is either congruent to or supplementary to $\angle FDH$ and $\angle FEB$ is either congruent to or supplementary to $\angle FEK.$ It follows that $\angle FDB$ and $\angle FEB$ are either congruent or supplementary in Case 2.
This concludes the proof that $\angle FDB$ and $\angle FEB$ are either congruent or supplementary angles. Now consider the cases "congruent" and "supplementary" individually.
Suppose $\angle FDB \cong \angle FEB.$ Since $F$ is on the bisector of $\angle DBE,$ it follows that $\triangle BDF \cong \triangle BEF$ with $BD \cong BE.$ But also $\angle CEF \cong \angle ADF$ (since each is supplementary to $\angle FDB$ or $\angle FEB$) and $\angle CFE \cong \angle AFD$ (by vertical angles), and together with $FD \cong FE$ we then have $\triangle ADF \cong \triangle CEF$ with $AD \cong CE.$ Since $D$ is between $A$ and $B$ and since $E$ is between $B$ and $C,$ it follows that $AB \cong BC,$ that is, $\triangle ABC$ is isosceles.
On the other hand, suppose $\angle FDB$ and $\angle FEB$ are supplementary angles. Let $\angle BAC = \alpha$ and $\angle ACB = \gamma.$ Then, since an exterior angle of triangle $\triangle ACE$ or $\triangle BCD$ is the sum of the other two interior angles, $\angle FDB = \alpha+\gamma/2$ and $\angle FEB = \alpha/2+\gamma.$ But then, since $\angle FDB$ and $\angle FEB$ are supplementary, $$ (\alpha+\gamma/2) + (\alpha/2+\gamma) = 180^\circ, $$ which implies that $\alpha+\gamma = 120^\circ.$ Hence $\angle ABC = 60^\circ.$
Therefore either $\triangle ABC$ is isosceles, or $\angle ABC = 60^\circ.$
As an aside, if the circle is tangent to $AB$ at $D,$ then it follows both that $\triangle ABC$ is isosceles and that $\angle ABC = 60^\circ,$ that is, $\triangle ABC$ is equilateral.
But if the circle is not tangent to $AB$ at $D,$ then $\triangle FHD$ and $\triangle FKE$ are not degenerate. Either $\triangle FHD$ and $\triangle FKE$ have the same orientation, in which case $\angle ABC = 60^\circ,$ or $\triangle FHD$ and $\triangle FKE$ have opposite orientations (are mirror images), in which case $\triangle ABC$ is isosceles. In either case it is not possible both for $\triangle ABC$ to be isosceles and for $\angle ABC$ to be $60^\circ.$
Partial solution: If $ABC$ is isosceles, then $FE$ is equal to radius of the circle which also means that $FD$ is a radius and $FD$ is perpendicular to $AB$ (radius to tangent). Thus $DBC$ is a right triangle and $\angle DBC=2 \angle DCB$. From here we conclude that $\angle DCB =30°$ and $\triangle ABC$ is equilateral.
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?
