For reference: In the regular $ABCDE$ pentagon, the diagonals $AC$ and $BE$ intersect at $P$. Calculate $PD$; if: $PC = 4m$. (Answer: $2\sqrt{10-2\sqrt5}m$)
My progress:
$PC = l_5=\frac{R}{2}\sqrt{10-2\sqrt5}=4 \implies
R\sqrt{10-2\sqrt5}=8\therefore R =\frac{8}{\sqrt{10-2\sqrt5}} \\
a_5 = \frac{R}{4}(\sqrt5+1) \\
\triangle BGP(36^o-54^o)\implies GP = (\sqrt5+1)k\\
\frac{l_5}{2}=(10-2\sqrt5)k \implies
k=\frac{2}{10-2\sqrt5}=\frac{1}{5-\sqrt5}=\frac{5+\sqrt5}{20} \implies\\ GP = {(1+\sqrt5)}.\frac{5+\sqrt5}{20}=\frac{3\sqrt5+5}{10}\\
DG=R+a_5=\frac{8}{\sqrt10-2\sqrt5}+\frac{R}{4}(\sqrt5+1)=\frac{8}{\sqrt10-2\sqrt5}+2(\sqrt5+1)\\
DG=\frac{2\sqrt50-12+2\sqrt10-4\sqrt5}{\sqrt10-2\sqrt5}\\
PD=DG-GP...$
I think I took the wrong "path"
Assuming $PC = 4$
Notice that $BC = PC$ (as $\angle BPC = \angle PBC$). Similarly, $AE = PE = PC$. So, $PCDE$ is a rhombus and $PD \perp CE$. Say they intersect at $M$. Then by Pythagoras in $\triangle PEM$,
$ \displaystyle PD = 2 PM = 2 \sqrt {PE^2 - \frac{CE^2}{4}}$
We know that the diagonals of a convex regular pentagon are in the golden ratio to its sides.
That leads to $CE = \dfrac{1 + \sqrt 5}{2} \cdot PC = 2 (1 + \sqrt 5)$
$ \therefore PD = 2 \sqrt{16 - (6 + 2 \sqrt 5)} = 2 \sqrt {10 - 2 \sqrt 5}$