TITLE: Prove this $BF=AF+CF$
QUESTION [8 upvotes]: In $\Delta ABC$ such $AB<AC<BC$,and the point $D$ on $BC$,and the  point $E$ in the extended line $BA$,such $$BD=BE=AC$$
Let the Circumcircle of triangle $\Delta ABC$ is $\Gamma_{1}$,and the circumcircle if triangle $\Delta BDE$ is $\Gamma_{2}$,if $\Gamma_{1}\bigcap \Gamma_{2}=F$
show that
$$BF=AF+CF$$

REPLY [3 votes]: Note that $\angle FED=\angle FBD=\angle FBC=\angle FAC$, and similarly $\angle FDE=\angle FCA$. It follows that $\triangle FED\sim\triangle FAC$. This implies that $FE=FA(\frac{DE}{AC})$ and $FD=FC(\frac{DE}{AC})$. Finally, according to Ptolemy's Theorem on cyclic quadrilateral $FEDB$, we see that:
$$\begin{align*}
(FB)(DE)&=(FE)(BD)+(FD)(BE)
\\ &=(FE)(AC)+(FD)(AC)
\\ &= (FA)(DE)+(FC)(DE)
\end{align*}
$$
from which $FB=FA+FC$ follows.