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For any two unequal even-sided regular polygons, the circumcircle around them bisects the segment connecting the vertices of the two polygons.

Here are some images illustrating the question:

For any two even-sided regular polygons with equal number of sides, I observed that the following relationship always hold:

$$CO=C_1O$$

Just for clarification, we can construct a circle with three points. In the above cases, the three circumcircles are formed by three-point pairs $(D,A,D_1),~(E,F,E_1,),~(F,G,F_1)$ respectively. Point $O$ is where the circumcircle intercepts the segment connecting the vertices of the polygons.
I have stumbled on this problem for a while, and couldn't figure out how to prove it. Any hints would be appreciated.

As I said in my comment, you can replace your polygons by circles. The picture is then as following:

You have two circles, tangent to each other at some point $T$, and two points (here, $P_1$ and $P_2$), one on each circle, such that the angle $widehat{TC_1P_1}$ and $widehat{TC_2P_2}$ are equal.

It is known that the center of the circle going through $T$, $P_1$ and $P_2$ is the
intersection of the perpendicular bissecors of $TP_1$ and $TP_2$, call this intersection I.

If one continues the lines $C_1P_1$ and $C_2P_2$ untill they intersect, it forms an isoceles triangle. Since the perpendicular bissectors or $TP_1$ and $TP_2$ are also the bissectors of the angles $widehat{TC_1P_1}$ and $widehat{TC_2P_2}$, the line going from the third vertex to $I$ happens to be the perpendicular bissector of $C_1C_2$. Hence since $C_1C_2$ is perpendicular to this diameter of the circle, $M$ must be the symetric of $T$ by the axis of this bissector.

Now it only remains to show that, with this information, $AM = MB$. Let's call U the middle of $C_1C_2$. Since $MU = UT$, $C_1U = UC_2$ becomes $C_1M + MU = UT + TC_2$, one can add the length of $AC_1$ on both side to get $AC_1 + C_1M + MU = UT + TC_2 + AC_1$. But $AC_1 = C_1T$ by definition of $C_1$, so $AC_1 + C_1M + MU = UT + TC_2 + C_1T$, which simplifies to $AM + MU = UT + C_1C_2$, and since $MU = UT$, to $AM = C_1C_2$, similarly, one can show that $MB = C_1C_2$. Hence $AM = MB$ as was to be shown.

This question can be answered by using the elementary complex cross-ratio. We are given a positive integer $n>1$ and define the $n$-th root of unity $z := e^{2pi i/n}$. Suppose we have two externally tangent circles in the complex plane. Without loss of generality, using rigid motions, the first circle is centered at the origin $P_0 := 0$ with radius $r_1$. The point of tangency being $P_1 := P_0 + r_1.$ The second circle is centered at $P_5 := P_1 + r_2$. Define the points $P_2 := P_0 + r_1 z$ on the 1st circle rotated clockwise from point $P_1$ and $P_3 := P_5 - r_2/z$ on the 2nd circle rotated counterclockwise from point $P_1$. Define
$P_4 := ((P_0-r_1) + (P_5+r_2))/2 = P_0 + r_2$ as the center of the circumcircle of the two tangent circles. This is labeled as $O$ in the diagrams.

The question is how to prove that the four points $P_1,P_2,P_3,P_4$ are concyclic. The answer is this is true if and only if their cross-ratio is real. Check that the cross ratio simplifies as
$$ frac{(P_1-P_3)(P_2-P_4)}{(P_2-P_3)(P_1-P_4)} =
frac{(r_2-r_2/z)(r_1 z-r_2)}{(r_1(z-1)-r_2(1-1/z))(r_1-r_2)} =
frac{-r_2}{r_1-r_2} $$
which proves the result. Notice that we did not need to use that $n$ is even and if $n=2$ then the four points are collinear. If $r_1=r_2$ then the result is still true since $P_1=P_4$ in that case so we have only three different points and any three non-collinear points are concyclic.

Here is a slightly more general setting:

Let $O_1,O_2$ be two points. Choose $C$ on the segment $O_1O_2$. The circles centered in $O_1,O_2$ through $C$ intersect the line $O_1O_2$ in ($C$ and) the (other) points $A_1$, and respectively $A_2$. Consider now from $A_1, A_2$ two rays in the same half plane w.r.t. $O_1O_2$ making the same angle with $O_1O_2$. Let $B_1,B_2$ be the intersection of the rays with the circles $(O_1)$, respectively $(O_2)$.

Then the points $C,B_1,B_2$ and the middle of the segment $A_1A_2$ are conciclic.

In the picture there is also a quick possible proof.

The center of the circle $(B_1CB_2)$, denoted by $M$, lies on the side bisectors of $B_1C$, respectively $B_2C$, being thus determined as their intersection. The centers $O_1$, respectively $O_2$ lie on them, and the triangle $MO_1O_2$ is isosceles, $MO_1=MO_2$, because of the base angles.

The perpendicular in $M$ on $O_1O_2$ is thus the mid point of the segment $O_1O_2$. Let $C^*$ be the symmetric of $C$ w.r.t. this perpendicular. Then $C^*$ is also on the circle $(B_1CB_2)$ centered in $M$, since $MC=MC^*$. It remains to observe that $C^*$ is the mid point of $A_1A_2$.

$square$