<s>[INST] Find $\frac{BF}{FC}$ in terms of $b,c$ If $AD$ is the bisector of angle $BAC$ and $E$ is the reflection of $D$ to point $M$(midpoint of $BC$).We construct point $F$  such that angles $BAF=EAC$.Then find $\frac{BF}{FC}$ in terms of $b,c$.

My attempt:We have:
$\frac{BF}{FC}=\frac{AB}{AC}*\frac{\sin BAF}{\sin CAF}$
But then what?
 [/INST] By the bisector theorem $BD=\frac{c}{b+c}a$ and $CD=\frac{b}{b+c}a$. 
Since $BM=CM=\frac{1}{2}a$, $BE=\frac{b}{b+c}a$ and  $CE=\frac{c}{b+c}a.$
If $\widehat{BAF}=\widehat{EAC}$, the lines $AE$ and $AF$ have to be symmetric with respect to $AD$.
The trilinear coordinates of $E$ are $[0,c^2,b^2]$, hence the trilinear coordinates of $F$ are $[0,b^2,c^2]$ and
$$\frac{BF}{FC}=\frac{c^2/b}{b^2/c}=\color{red}{\frac{c^3}{b^3}}.$$

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