Proving a proportion involving the diagonals of a regular pentagon 
Given regular pentagon $ABCDE$, prove that
  $$\frac{DA}{DK} = \frac{DK}{AK}$$

My attempt: By the Triangle Proportionality Theorem, 
$$\frac{AK}{KD} = \frac{EK}{KB}$$ 
I'm not too sure about where to go next. Perhaps $\triangle KED \sim \triangle KDB$ by Angle-Angle Similarity?

 A: Since it's a regular pentagon, each of the internal angles at $A, B, C, D, E$ have an angle of $108°$. Using that $DC = CB$, then $\triangle DCB$ is isosceles, with $\angle CDB = \angle CBD = 36°$. In fact, using isosceles triangles and sum of angles in a triangle, you get
$$\angle BDA = \angle ADE = \angle BEA = 36° \tag{1}\label{eq1}$$
and
$$\angle DEK = \angle DKE = \angle DAB = \angle DBA = \angle BKA = 72° \tag{2}\label{eq2}$$
Thus, $\triangle DBA \sim \triangle DKE$, so
$$\frac{DA}{DK} = \frac{AB}{EK} \tag{3}\label{eq3}$$
Since $\triangle DKE \cong \triangle BKA$ (by all angles being equal and $DE = AB$), then $EK = AK$. Also, $AB = DE$, due to it being a regular pentagon, and $DE = DK$ due to $\triangle DKE$ being isosceles, so $AB = DK$. Thus, the RHS of \eqref{eq3} can be rewritten as
$$\frac{DA}{DK} = \frac{DK}{AK} \tag{4}\label{eq4}$$
which is what was requested to be proven.
A: Since $$\measuredangle DEK=\measuredangle DKE,$$ we obtain $DK=DE=AB$ and it's enough to prove that
$$AB^2=AK\cdot AD,$$ for which it's enough to prove that $AB$ is a tangent to the circumcircle of $\Delta DKB,$ for which it's enough to prove that
$$\measuredangle KBA=\measuredangle KDB,$$ which is obvious.
