Question: In the $\triangle ABC$ triangle, where the angle $\angle ~A$ is 120 degrees, the $AF$, $BE$ and $CD$ bisectors intersect at point $I$. Demonstrate that $BD + CE + 2IF = BC$.

I was thinking of applying the bisector theorem, but it seems I'm not getting anywhere with it. Then, I thought about Menelaus, but again, I failed. How should I look at this problem? I just really need a hint.

Answer: 
[![enter image description here](https://i.stack.imgur.com/ViNQg.png)](https://i.stack.imgur.com/ViNQg.png)

Let $DT\perp BI $ such that $DT$ touches $BC$ at $T$ and $EP\perp CI$ such that $EP$ touches $BC$ at $P$. Then, because $BI$ and $CI$ are angle bicectors, we have $\triangle BDI\cong \triangle BTI$ and $\triangle CEI\cong \triangle CPI$. Hence, $$\angle TIF = \angle BIF-\angle BIT = (\angle BAF+\angle ABI)-\angle BIT = (\angle BAF+\angle ABI)-\angle BID$$
Here, $\angle BAF = 60^\circ$, $\angle BID = \angle IBC +\angle ICB = 90^\circ-\frac{1}{2}\angle BAC = 30^\circ$, and $\angle BIT = \angle BID$ by congruency, so $$\angle TIF = 30^\circ +\angle ABI = 30^\circ + \angle TBI = \angle TBI + \angle TIB = \angle ITF$$
We can derive $\angle FPI = \angle FIP$ from exactly the same procedures, except interchanging the letters $(ADBT)$ with $(AECP)$. This implies that $\triangle ITP$ is a right triangle, and $F$ is the midpoint of $TP$. 

Hence, $$BC = BT+PT+CP = BD + 2IF + CE$$ as desired.
A different solution (less efficient than Lazy Lee's) can employ the bisector theorem.  

By denoting $AB,AC,BC$ through $c,b,a$ as usual, we have:
$$ BD=\frac{ac}{a+b},\quad CE=\frac{ab}{a+c} $$
[By Stewart's theorem](https://proofwiki.org/wiki/Length_of_Angle_Bisector) the length of the angle bisector through $A$ is provided by
$$ AF^2 = \frac{bc}{(b+c)^2}\left((b+c)^2-a^2\right) $$
but since $\widehat{A}=120^\circ$ we also have $a^2=b^2+c^2+bc$ by the cosine theorem, hence, simply,
$$ AF = \frac{bc}{b+c} $$
and by Van Obel's theorem we have $\frac{AI}{IF}=\frac{b+c}{a}$, hence $IF=\frac{abc}{(b+c)(a+b+c)}$.  
 The claim boils down to proving that
$$ \frac{ac}{a+b}+\frac{ab}{a+c}+\frac{2abc}{(b+c)(a+b+c)} = a $$
and such identity is straightforward to prove by exploiting $a^2=b^2+c^2+bc$ again.