Conditions of inscribing a hexagon in a circle What are the conditions that when they are met by a hexagon, enable us to assume it is possible to inscribe it in a circle?
Edit: Is it possible to find conditions without coordinates?
 A: If $AB$ is a side of the hexagon and $CDEF$ its other vertices, then it can be inscribed in a circle if
$$
\angle ACB=\angle ADB=\angle AEB=\angle AFB.
$$
A: Let $ABCDEF$ be a convex hexagon.
Given a side, let's say $AF$, if angles $ABF,ACF,ADF,AEF$ are equal then the hexagon is inscrivible in a circle.
Hope this helps
A: Here is an algebraic solution: let's call the coordinates of the vertices $x_i$ and $y_i$, with $i=\overline{1,6}$. If these are points on a circle, then the center of the circle is at $(x,y)$, and the radius is $r$. We write the equation for the length of the radius for each vertex:
$$(x_i-x)^2+(y_i-y)^2=r^2$$
Using $i=1$ and $i=2$, we subtract the first equation from the second:$$x_1^2-x_2^2-2x(x_1-x_2)+y_1^2-y_2^2-2y(y_1-y_2)=0$$
or
$$-2x(x_1-x_2)-2y(y_1-y_2)+(x_1^2+y_1^2-x_2^2-y_2^2)=0$$
You can write similar equations for $i=3,4$ and $i=5,6$. You have a system of three equations, with only two unknowns ($r$ does not appear here). This is consistent if and only if the determinant is zero.
$$\begin{vmatrix}
x_1-x_2 & y_1-y_2 & x_1^2+y_1^2-x_2^2-y_2^2 \\ 
x_3-x_4 & y_3-y_4 & x_3^2+y_3^2-x_4^2-y_4^2 \\ 
x_5-x_6 & y_5-y_6 & x_5^2+y_5^2-x_6^2-y_6^2
\end{vmatrix}=0$$
At the same time you want to have a solution, so 
$$\begin{vmatrix}
x_1-x_2 & y_1-y_2  \\ 
x_3-x_4 & y_3-y_4 
\end{vmatrix}\ne0$$
A: The algorithm is routine, but tedious, so best implemented as a program . . .

Suppose you have the $6$ consecutive side lengths, and for each pair of consecutive sides, you know the measure of the interior angle between them.

Let $A,B,C,D,E,F$ denote the $6$ consecutive vertices.

Since you know the lengths $AB,BC$, and the measure of angle $ABC$, you can fully solve triangle $ABC$, using the law of cosines.

Then use the generalized law of sines to compute the circumradius $R$ of triangle $ABC$.

Do the same for triangles $BCD,\;CDE,\;DEF$.

If all the circumradii are equal, hexagon $ABCDEF$ is cyclic, else it's not.

Based on the above algorithm, you only need $5$ consecutive side lengths, and the measures of the $4$ interior angles between consecutive pairs of those sides.

Note: Just knowing all $6$ side lengths or all $6$ interior angle measures is definitely not sufficient. You need a combination.

Also note: The algorithm generalizes directly to the the case of an $n$-gon.
