<s>[INST] Find the $\frac{k}{8}$ when the rest of the lengths are given 6 points $A,B,C,D,E,F$ are on a circle in this order and line segments $AD,BE,CF$ intersect at one point. If $AB=1,BC=2,CD=3,DE=4,EF=5,FA=\frac{k}{8}$ find the value of $k$.
I tried to sum the total lengths of the arc:
$1+2+3+4+5+\frac{k}{8}=2\pi r$
$15+\frac{k}{8}=2\pi r$
$\frac{120+k}{8}=2\pi r$
$\frac{120\cdot 7+k}{8\cdot 2\cdot 22}=r$
$\frac{840+k}{352}=r$
So I tried substituting the values in $2\pi r$
$2\pi \frac{840+k}{352}=\frac{120+k}{8}$
$\frac{36,960+44k}{2464}=\frac{120+k}{8}$
$295,680+352k=295,680+44k$
$352k=44k$
And if I cut $k$ from both the side, $352=44$ would remain which is absurd.
What went wrong here? And how can I get the value of $k$?
If I missed something obvious, please be gentle
 [/INST] I think they are chord lengths and not arc lengths.

By Intersecting Chords theorem or by Inscribed Angle theorem, $\triangle GAB \sim \triangle GED$
So, $GD = 4y, GE = 4x$
Similarly, $\triangle GBC \sim \triangle GFE$
So, $GE = \frac{5z}{2}, GF = \frac{5y}{2}$
$\triangle GCD \sim \triangle GAF$
So, $GF = \frac{5y}{2} = \frac{k}{24} \cdot GD = \frac{ky}{6} \implies k = 15$
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