Neem AB=x en noem de hoek $\alpha$. Er geldt:

$
\eqalign{
  & \tan \left( \alpha  \right) = \frac{3}
{x} \Rightarrow   \cr
  & \tan \left( {2\alpha } \right) = \frac{{2\tan \left( \alpha  \right)}}
{{1 - \tan ^2 \left( \alpha  \right)}}  \cr
  & \tan \left( {2\alpha } \right) = \frac{{2 \cdot \frac{3}
{x}}}
{{1 - \left( {\frac{3}
{x}} \right)^2 }} = \frac{{\frac{6}
{x}}}
{{1 - \frac{9}
{{x^2 }}}}  \cr
  & \tan \left( {2\alpha } \right) = \frac{x}
{5}  \cr
  & Dus:  \cr
  & \frac{{\frac{6}
{x}}}
{{1 - \frac{9}
{{x^2 }}}} = \frac{x}
{5} \Rightarrow x^2  = 39  \cr
  & AF^2  = x^2  + 5^2  = 39 + 25 = 64  \cr
  & AF = 8 \cr}
$

 @EJMenzel