$F$ is the incentre of $\Delta ABC$. $AF$ is joined to meet $BC$ at $D$ and the circumcircle of $\Delta ABC$ at $E$. If $ED = 4cm$ and $DF = 3cm$, find $8AI$.
MY WORK :- Regarding this question I could think of only 2 formulas : ${AF\over FD} = {AC+AB\over BC}$ and $AB*AC = BD*DC + AD^2$ however none of them are helping in finding the solution. I feel as if the solution is revolving around these formulas, but still I am not able to reach to a fruitful end.
We use Fact 5 that $E$ is center of circle passing through $B,F,C,J$ where is $J$ is $A$-excenter. Thus $$EF=7=EB=EC=EJ$$
Now by power of point of $D$ wrt these two circles
$$AD\cdot DE =BD\cdot CD= DF\cdot DJ$$
$$\Rightarrow (AF+3)\cdot 4 = 3\cdot 11$$
$$\Rightarrow 8AF = 42$$