Problem about the similarity between triangles

Question

Referring to the following image:

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 1

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!

Answer 2

HINT: draw seg BD and CD and use angles in alternate segment theorem

Answer 3

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 $$AB\cdot AE = AD^2 = AC\cdot AF$$