Question: Referring to the following image:

[![enter image description here](https://i.stack.imgur.com/Q3UF3.png)](https://i.stack.imgur.com/Q3UF3.png)

how can it be demonstrated that $AB : AC = AF : AE$ ?

The only theorem that seems to me to be useful is the *theorem of the secante and of the tangent*: $AE : DE = DE : BE $ or equally: $AF : DF = DF : CF$, but I can not conclude anything. 

Some idea?

Answer: 
HINT: draw seg BD and CD and use angles in alternate segment theorem
If you are familiar with the inversion, then perform it at $A$ and $r=AD$. Then the circle goes to tangent at $D$ and so $D\mapsto D$, $B\mapsto E$ and $C\mapsto F$. So we have $$AB\cdot AE = AD^2 = AC\cdot AF$$