ABCDE is a regular pentagon. $\angle AFD = \angle EKC$
$|FH|=1$ cm; $|AH|=3$ cm
What is $|DK|?$
I know that triangles $EFA$ and $DEK$ are similar and that $|EK|=4$ cm. Also because this is a regular pentagon each one of the interior angles are $108^o$. Naming similar angles inside the pentagon, I tried to find an isosceles triangle, but I couldn't. I can't progress any further from here.
How can I solve this problem?