Finding $|HD|$ of a cyclic triangle $\triangle BCD$ where two points $E$ and $F$ are on circumference and $CD \parallel EF$ 
In $\triangle BCD$, $BC$ = $5$ and $G$ & $H$ are two points on $CD$ such that $CG$ = 1, $GH$ = 3 and $HD$ = 2. $BG$ and $BH$ intersect the circumcircle of $\triangle BCD$ at the point $E$ and $F$ respectively. If $EF$ $\parallel$ $CD$ then what is the length of $BD$?


Whatever I thought is that in the event of $EF$ $\parallel$ $CD$, we can show that $\triangle BGH$ $\sim$ $\triangle BEF$. But how can I use of the rest property for solving this problem? That was too hard for me.
 A: 
Connect $CE, DF$, let $\angle GCE=\angle HDF=x$
$${BD\over BC}={sin(\angle BCD)\over sin(\angle BDC)} = {sin(\angle HFD) \over sin(\angle GEC)} = {sin(\angle HFD) \over x}\times{x\over sin(\angle GEC)}=$$
$$={2\over HF}\times{GE\over 1} = 2\times{GE\over HF} = 2\times{BG\over BH} = 2\times\sqrt{BG\times GE\over BH\times HF} = 2\times \sqrt{1\times 5\over 4\times 2}= \sqrt{5\over 2}$$
Therefore $BD=5\sqrt{5\over 2}$
A: Let $BD=x$, $GE=y$ and $HF=z$.
Thus, by similarity we obtain:
$$\frac{ED}{5}=\frac{y}{1},$$ which gives
$$CF=ED=5y.$$
Also, $$\frac{DF}{5}=\frac{z}{4},$$ which gives
$$CE=DF=\frac{5z}{4}.$$
Now, $$BG\cdot y=1\cdot5,$$ which gives $$BG=\frac{5}{y}.$$
Also, $$BH\cdot z=4\cdot2,$$ which gives $$BH=\frac{8}{z}.$$
We have also $$\frac{BD}{CF}=\frac{HD}{HF}$$ or
$$\frac{x}{5y}=\frac{2}{z}.$$
Now, by Ptolemy for $BCED$ and for $BCFD$ we obtain:
$$5\cdot5y+\frac{5z}{4}\cdot x=6\left(y+\frac{5}{y}\right)$$ and 
$$5\cdot\frac{5z}{4}+x\cdot5y=6\left(z+\frac{8}{z}\right).$$
We got the system of three equations with three variables.
Can you end it now?
I got $x=\sqrt{62.5}.$
