What's the measure of the segment HD in the pentagon below?

For reference:If from a point H of the circumference circumscribed to a regular pentagon ABCDE segments are traced to the five vertices A, B, C, D and E; if: HE $\in \overset{\LARGE{\frown}}{AE}$; HA + HE + HC = 19 and HB=9. Calculate HD. (Answer:$10$)

My progress:

$HA+HE+HC = 19\\HDCB: DH.BC+BH.CD=BD.CH\implies\\ DH.l_5+9.l_5=BD.CH\\ {l_5(DH+9)=BD.CH}\\ BD=l_5\frac{(\sqrt5+1)}{2} \implies \boxed{DH+9=CH\frac{(\sqrt 5+1)}{2}}\\ AHCB: AH.l_5+CH.l_5=AC.9\\ l_5(AH+CH)=l_5\frac{(\sqrt5+1)}{2}.9\implies\\ \boxed{AH+CH=9.\frac{(\sqrt5+1)}{2})\\ }\\HEDC: HE.l_5+l_5.CH=CE.DH\implies\\ l_5(HE+CH) = l_5\frac{(\sqrt5+1)}{2}.DH\implies\\ \boxed{HE+CH =DH.\frac{\sqrt5+1)}{2}}$

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Solution 1:

Let $x$ denote the side-length of the pentagon. We know that diagonals of a regular pentagon are in the golden ratio to its side. Therefore a diagonal is $\varphi x$.

Now apply Ptolemy's theorem to following quadrilaterals.

$\begin{align}\boxed{ABCH}\to &HA\cdot x+HC\cdot x=HB\cdot \varphi x\\ \to &HA+HC=HB\cdot \varphi\tag1\end{align}$

$\boxed{EDCH}\to HE+HC=HD\cdot \varphi\tag2$

$\boxed{HBCD}\to HB+HD=HC\cdot \varphi\tag3$

$\begin{align}\boxed{AHED}\to &HA+HE\cdot\varphi=HD\\ \to &HD-HA=HE\cdot\varphi\tag4\end{align}$

$\boxed{BAHE}\to HB-HE=HA\cdot\varphi\tag5$

Now add them all recalling $HA+HE+HC=19$ and $HB=9.$

Thus we have $$2(HB+HC+HD)=(28+HD)\varphi$$

From $(3)$, $HC=\dfrac{(HB+HD)}{\varphi}=(9+HD)(\varphi-1).$

$\small \left(\because \dfrac1{\varphi}=\varphi-1\right).$

Now substitute this in your equation and simplifying, you will get your answer.