Decide diagonals on a hexagon inscribed in a circle I have been struggling a bit with a math-task.
The question is as follows:
The hexagon ABCDEF is inscribed in a circle. The length of AF is 31 while the other five sides has the length of 81. Tell the sums of the diagonals from A. 
Dividing the hexagon into a quadrilateral AFEB and using Ptolemaios theorem gives $AE\cdot BF=AF\cdot BE+AB\cdot EF=31\cdot BE+81^2$
Dividing the hexagon into a quadrilateral AFED and using Ptolemaios theorem gives $AE\cdot DF=AF\cdot DE+AD\cdot EF=31\cdot81+81\cdot AD=81(31+AD)$
I just keep getting more unknown variables and I really don't know how to proceed. Please help, I hope my effort has been explained clearly enough. Also, please give tips on how to structure and declare my questions and explainations better.

$\hspace1in$ 
Label the sides and diagonals of the hexagon as shown.


*

*$a = AF = 31$

*$b = AB = BC = CD = DE = EF = 81$

*$c = AC = BD = DF$

*$d = AD = CF$

*$e = AE$.


Apply Ptolemy's theorem to quadrilaterals $ABCD$, $ACDF$ and $ADEF$, we obtain
$$\begin{align}
c^2 &= bd + b^2\\
d^2 &= c^2 + ab\\
ce &= bd + ab
\end{align}$$
The first two equations tell us
$$d^2 = b(a+b+d) \iff d^2 = 81(d+112) \iff (d-144)(d+63) = 0 \implies  d = 144$$
Substitute this back into first equation, we obtain
$$c = \sqrt{b(b+d)} = \sqrt{81(81+144)} = 135$$
Throw this into last equation, we find
$$e = \frac{b(a+d)}{c} = \frac{81(31+144)}{135} = 105$$
The desired sum of diagonals from $A$ equals to
$$AC + AD + AE = c + d + e = 135 + 144 + 105 = 384$$
