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: 
Because $$\measuredangle ACB=\measuredangle BDA=90^{\circ}-\measuredangle BDE=\measuredangle BED,$$
which says that $\Delta ABC\sim \Delta AFE$, which gives
$$\frac{AB}{AF}=\frac{AC}{AE}$$ or
$$AB:AC=AF:AE$$ abd we are done!
HINT: draw seg BD and CD and use angles in alternate segment theorem