http://www.mathlinks.ro/viewtopic.php?t=313492

Let $ K\in AB$ such that $ \angle BKC = \angle ABI$ (A is between B and K). I will prove that $ \frac {BF}{FK} = \frac {1}{2}$. $ *120^{o} = \angle B + \angle C$.$ \Longrightarrow 60^{o} = \frac {\angle B}{2} = \frac {\angle C}{2}$. $ \Longrightarrow\angle ACK = \angle ICA$. Draw $ AT\perp AK ( T\in KC)$ and $ IM\perp BA(M \in BA)$ We have $ \angle TAC = \angle CAI = 30^{o}$ $ \Longrightarrow \bigtriangleup IAC = \bigtriangleup TAC$. $ \Longrightarrow TA = AI = 2IM$ on the other hand $ \bigtriangleup KAT \thicksim \bigtriangleup BIK$ $ \Longrightarrow \frac {BM}{AK} = \frac {IM}{AT} = \frac {1}{2}$. Notice that $ \frac {MF}{FA} = \frac {MF}{FI} = \frac {1}{2}$. Hence $ \frac {BF}{FK} = \frac {BM + MF}{FA + AK} = \frac {1}{2}$. q.e.d

Let $Q$ be the foot of the perpendicular from $P$ to $AB$, and let $R$ be on $\overline{AQ}$ so that $QR=QB$. We have $PR=PB$, so $\angle FBI=\frac{1}{2}B=\frac{1}{2}\angle BRP=\frac{1}{2}(\angle BFP+\angle RPF)$. Hence, it suffices to show that $\angle BFP=\angle RPF$, i.e. $RP=RF$. We have $BC=BI\cos\frac{B}{2}+CI\cos\frac{C}{2}$. But $CI\sin\frac{C}{2}=BI\sin\frac{B}{2}$, so \begin{align*}BC&=BI\frac{\cos\frac{B}{2}\sin\frac{C}{2}+\sin\frac{B}{2}\cos\frac{C}{2}}{\sin\frac{C}{2}} \\ &=BI\frac{\sin (\frac{B}{2}+\frac{C}{2})}{\sin\frac{C}{2}} \\ &=BI\frac{\sin 60^{\circ}}{\sin\frac{C}{2}}\\ \end{align*} Hence $BI=2BC\sin\frac{C}{2}\frac{1}{\sqrt{3}}$. By the law of sines on $\triangle BFI$, we have \[\frac{BF}{BI}=\frac{\sin (60^{\circ}+\frac{B}{2})}{\sin 60^{\circ}},\] so \begin{align*}BF&=\frac{2}{\sqrt{3}}BI\sin (60^{\circ}+\frac{B}{2})\\ &=\frac{4}{3}BC\sin\frac{C}{2}\sin (60^{\circ}+\frac{B}{2})\\ \end{align*} However, $2\sin\frac{C}{2}\sin (60^{\circ}+\frac{B}{2})=\cos (A+\frac{B}{2}+\frac{C}{2})=\cos B-\cos 120^{\circ}=\cos B+\frac{1}{2}$. It follows that $BF=\frac{2}{3}BC\cos B+\frac{1}{3}BC=2PB\cos B+PB=BR+RP$. But $BF=BR+RF$, so $RP=RF$, as desired. $\Box$

it can be solved by use of intersection point of $FI$ and $BC$(name it $D$) and of congruency of $BFD$,$ABC$ with ratio:$\frac{a(a+c)}{2p}$ and computing the $\frac{BP}{PD}$ with sines,...

Notations:$IB \cap FP=K,IB \cap AC=L,\angle{BFP}=\theta$ Proof$IF \parallel AL \Rightarrow \frac{FB}{c}=\frac{IB}{BL}=\frac{a+c}{a+b+c} \Rightarrow \boxed{BF=\frac{(a+c)c}{a+b+c}}$ We also see that $BP=\frac{a}{3}$ and $b^2+c^2-a^2=bc$...(1)(From the fact that $\angle{A}=60^{\circ}$) Now we start our manipulations: $\theta=\frac{B}{2}$ $\Leftrightarrow \frac{sin(B+\theta)}{sin\theta}=\frac{sin\frac{3B}{2}}{sin\frac{B}{2}}=3-4sin^2\frac{B}{2}=1+2cosB=\frac{a^2+c^2-b^2+ac}{ac}$ $\Leftrightarrow \frac{3ac+c^2}{a^2+ab+ac}=\frac{a^2+c^2+ac-b^2}{ac}$(The left side follows by applying sine rule in $\triangle{BPF}$) $\Leftrightarrow 3a^2c^2+3ac^3=(a^2+c^2+ac-b^2)(a^2+ab+ac)=(b^2+c^2-bc+c^2+ac-b^2)(a+b+c)$(Substituting from (1)) $\Leftrightarrow 3c(a+c)=(2c+a-b)(a+b+c)=2ac+2bc+2c^2+a^2+ab+ac-ab-b^2-bc$ $\Leftrightarrow c^2=bc+a^2-b^2$ $\Leftrightarrow b^2+c^2-a^2=bc$ $\Leftrightarrow \angle{A}=60^{\circ}$ which is true...

Let $\angle{BFP}=x$ So, $\frac{BF}{BP}=\frac{sin(x+\angle{B})}{sinx}$ $2r=AI=\sqrt{3}AF=4Rsin(\frac{\angle{B}}{2})sin(\frac{\angle{C}}{2})$ So we obtain, $2\sqrt{3}sin(\angle{C})-4sin(\frac{\angle{B}}{2})sin(\frac{\angle{C}}{2})=\frac{sin(x+\angle{B})}{sinx}$ $\implies 2\sqrt{3}sin(\frac{\pi}{3}+\angle{B})-4sin(\frac{\angle{B}}{2})cos(\frac{\pi}{6}+\frac{\angle{C}}{2})=\frac{sin(x+\angle{B})}{sinx}$ $\implies 3cos(\angle{B})+2sin^2(\frac{\angle{B}}{2})=\frac{sin(x+\angle{B})}{sinx}$ $\implies 3-4sin^2(\frac{\angle{B}}{2})=\frac{sin(x+\angle{B})}{sinx}$ $\implies \frac{sin(\frac{\angle{B}}{2}+\angle{B})}{sin(\frac{\angle{B}}{2})}=\frac{sin(x+\angle{B})}{sinx}$ $\implies cot(\frac{\angle{B}}{2})=cotx$ $\implies x=\frac{\angle{B}}{2}$

Synthetic proof is not at all difficult if we proceed along right lines. Let $P'$ be on $BC$ s.t. $\angle{BFP'}=B/2$.Similarly take $E$ on $AC$ where $IE$ is parallel to $AB$,and $Q$ on $BC$ with $\angle{CEQ}=C/2$.Let $X$ be the intersection of $QE$ and $AB$. It is easy to see that $BFI$ is congruent to $XFE$.So $BF=FX$.Also we can easily show that $FP'$ is parallel to $EQ$.So $P'$ is the midpt of $BQ$.Similarly $Q$ is the midpt of $P'C$.So $P=P'$ and we are done .

more solutions here edit, it's a pity old links not to work, e.g. livetolove212 wrote: http://www.mathlinks.ro/viewtopic.php?t=313492