I want to calculate the area of the $2n$ polygon outside
I already know the side length of the $2n$-polygon which is within the unit circle I have called the lengt $l_{2n}$, the length for the $n$ polygon outside I have defined as $L_n$. I could calculate the lengt of the triangle for the $l_{2n}$ polygon. I am trying to calculate the length which I have marked with $?$. How can I do it ?
I have added Pictures for the case $n=4$. But can someone also explain me why the calculations for this case would be also true for the General case?
I also realized that for this case $n=4$ one can arrange the outer $n$ polygon in two different ways. Why must this also be true for the General case?
I used GeoGebra online and snipping tool for the Pictures
I am Looking for a solution which makes use of congruency
This is not a solution making use of congruency, at least not directly. Everything said here is derived (by a pretty long chain of reasoning) from congruence arguments, but it may be too far removed to make you happy. So be it.
Let's look at the case $n = 4$. For a unit circle, the points of tangency will be $$ P_i = (\cos \frac{2\pi i}{4}, \sin \frac{2\pi i}{4} ), i = 0, 1, 2, 3 $$ (That's for the square with vertically-aligned sides). By replacing the "4" in the denominator with $k$, you get the coordinates for the tangent points of the $k$-gon externally tangent to the circle, with $i$ going from $0$ to $k-1$.)
The tangent vector to the circle at the point $P_i$ is $$ v_i = (-\sin \frac{2\pi i}{4}, \cos \frac{2\pi i}{4} ), i = 0, 1, 2, 3. $$ The normal vector is just $$ n_i = (\cos \frac{2\pi i}{4}, \sin \frac{2\pi i}{4} ), i = 0, 1, 2, 3 $$ (i.e., it has the same coordinates at $P_i$.
The equation of the line $L_i$ containing the side of the square that's tangent at $P_i$ is therefore $$ ((x, y) - P_i) \cdot n_i = 0, $$ or, in coordinates,
$$ (x - \cos \frac{2\pi i}{4}, y -\sin \frac{2\pi i}{4} ) \cdot (\cos \frac{2\pi i}{4}, \sin \frac{2\pi i}{4} ) = 0 $$ That's getting cumbersome to read, so let's call the coordinates of $P_i$ by the names $x_i$ and $y_i$, so that we have \begin{align} x_i &= \cos \frac{2\pi}{4},\\ y_i &= \sin \frac{2\pi}{4}, i = 0, 1, 2, 3 \end{align}
That means that the equation for $L_i$ becomes $$ (x-x_i, y-y_i) \cdot (x_i, y_i) = 0 $$ or, writing it out, $$ x_i x + y_i y = x_i^2 + y_i^2 $$ But because $P_i = (x_i, y_i)$ is on the unit circle, we have $x_i^2 + y_i^2 = 1$. So the equation of $l_i$ becomes $$ x_i x + y_i y = 1. $$
Now let's find the coordinates of the corners of the polygon. The corner that lies between $P_i$ and $P_{i+1}$ --- let's call it $Q_i$ --- must lie on both tangent lines, i.e., it must have coordinates $(x,y)$ that satisfy two equations: \begin{align} x_i x + y_i y &= 1\\ x_{i+1} x + y_{i+1} y &= 1\\ \end{align}
Let's figure out the coordinates for the particular case where $i = 0$, so we're looking at \begin{align} x_0 x + y_0 y &= 1\\ x_1 x + y_{1} y &= 1\\ \end{align} where, as it happens, $x_0 = 1$ and $y_0 = 0$. The equations reduce to \begin{align} x &= 1\\ x_1 x + y_{1} y &= 1\\ \end{align} which, when we plug $x = 1$ from the first equation into the second equation, gives us \begin{align} x &= 1\\ x_1 + y_{1} y &= 1\\ y_{1} y &= 1-x_1\\ y &= \frac{1-x_1}{y_1}\\ \end{align} Now, letting $\alpha_i = \frac{2\pi}{i}$ denote the angle for $P_i$, this gives us the point where $$ x = 1, y = \frac{1 - \cos \alpha_1}{\sin \alpha_1}. $$
Whew! The cool thing is that this formula works whether we're looking at $n = 4$ (the square) or $n = 10$ (the decagon) or any other positive integer $n$. You just have to use the formula $$ \alpha_i = \frac{2 \pi i}{n} $$ instead of $$ \alpha_i = \frac{2 \pi i}{4} $$ to make it work/
In the case of the square, $\alpha_1 = \frac{2\pi}{4}$, so its sine and cosine are both $s = \frac{\sqrt{2}}{2}$, and we get \begin{align} x &= 1 & y &= \frac{1-x_1}{y_1}\\ x &= 1 & y &= \frac{1-s}{s}\\ x &= 1 & y &= \frac{1}{s} - 1\\ x &= 1 & y &= \frac{2}{\sqrt{2}} - 1\\ x &= 1 & y &= {\sqrt{2}} - 1. \end{align}
But in general, the formula $$ x = 1, y = \frac{1 - \cos \alpha_1}{\sin \alpha_1}. $$ works for the first point of the circumscribing $n$-gon for any $n$. Let's call that point $Q_1$, and name the coordinate $u_1, v_1$.
To find the other points, you could mess around with solving similar pairs of equations, but it's far easier to just rotate the point $Q_i$ around the origin by angles $\alpha_1, \alpha_2, \ldots$. The point $(s, t)$, rotated by angle $\alpha_i$, becomes the point $(s', t')$, where \begin{align} s' &= \cos (\alpha_i)s - \sin(\alpha_i) t \\ t' &= \sin (\alpha_i)s + \cos(\alpha_i) t. \end{align} Applying this to our particular point $Q_1 = (1, \frac{1 - \cos \alpha_1}{\sin \alpha_1})$, we get $$ Q_i = \biggl( \cos (\alpha_i) - \sin(\alpha_i)\frac{1 - \cos \alpha_1}{\sin \alpha_1}, \sin (\alpha_i) + \cos(\alpha_i)\frac{1 - \cos \alpha_1}{\sin \alpha_1}\biggr) $$ where $$ \alpha_i = \frac{2 \pi i}{n} $$ and those are the coordinates of the points of the externally-tangent $n$-gon.