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
$$BF=AF+CF$$
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.
