TITLE: In $\triangle ABC$, if angle bisectors $AE$ and $CD$ meet at incenter $F$, and $|FE|=|FD|$, then the triangle is isosceles or $\angle B=60^\circ$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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 $\hat A=\hat C$.
$\square$