Circle measurement of Archimedes Let $f_n$ or $F_n$ be areas of the regular $n-$ polygon described to the unit circle or circumscribed. Show 
$f_{2n}=\sqrt{f_nF_n}$ and $F_{2n}=\frac{2f_{2n}F_{n}}{f_{2n}+F_n}$
In the solutions there is only the fact that this follows from elementary geometric considerations.
I tried to explain the result by elementary geometry, but I did not succeed.
I have already searched the net for approaches. For example you can construct a $2n$ corner out of a $n-$ corner which is in a circle by drawing a point between two corners of a side.
To clarify what I mean I have added a sketch using the example of a 4 corner. 
Why is the geometric mean of the area content of the outer and inner 4 corners now the area content of the inner $8$ corner?
I have no approach to the formula for the outer $2n$ corner.
How can I draw the $2n$ corner from the outer $n$ and why is the harmonic mean of the area of the inner $2n$ corner and the area of the inner $n$ corner?
I hope someone can help me.


Let $l_n$ and $L_n$ be the lengths of the sides of the regular $n$-polygon inscribed in, or circumscribed to, the unit circle. The inscribed polygon with $n$ sides is formed by $n$ equal triangles, with base $l_n$ and altitude $\sqrt{1-l_n^2/4}$. We have then:
$$
f_n={n\over2}l_n\sqrt{1-l_n^2/4},\quad F_n={n\over2}L_n.
$$
On the other hand $L_n:l_n=1:\sqrt{1-l_n^2/4}$, that is:
$$
L_n={l_n\over\sqrt{1-l_n^2/4}},\quad\text{and}\quad 
F_n={n\over2}{l_n\over\sqrt{1-l_n^2/4}}.
$$
The area of the inscribed polygon with $2n$ sides is obtained by adding to $f_n$ the area of $2n$ small triangles:
$$
f_{2n}=f_n+2n\cdot{1\over2}{l_n\over2}\big(1-\sqrt{1-l_n^2/4}\big)=
{n\over2}l_n=\sqrt{f_n\cdot F_n}.
$$
For $F_{2n}$ one can do a similar reasoning:
$$
l_{2n}^2=l_n^2/4+\big(1-\sqrt{1-l_n^2/4}\big)^2=2\big(1-\sqrt{1-l_n^2/4}\big),
$$
$$
\tag{*}
F_{2n}={L_{2n}^2\over l_{2n}^2}f_{2n}={1\over1-l_{2n}^2/4}f_{2n}=
{2\over1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n
$$
and finally:
$$
{1\over f_{2n}}+{1\over F_{n}}={2\over nl_n}\big(1+\sqrt{1-l_n^2/4}\big)
={2\over F_{2n}}.
$$
EDIT.
To obtain the last result in $(*)$, we must substitute 
$l_{2n}^2=2\big(1-\sqrt{1-l_n^2/4}\big)$ and
$f_{2n}={n\over2}l_n$ into 
$F_{2n}=(1-l_{2n}^2/4)^{-1}f_{2n}$:
$$
F_{2n}=\left(1-{l_{2n}^2\over4}\right)^{-1}f_{2n}=
\left(1-{2\big(1-\sqrt{1-l_n^2/4}\big)\over4}\right)^{-1}{n\over2}l_n=
\left({1\over2}+{1\over2}\sqrt{1-l_n^2/4}\right)^{-1}{n\over2}l_n=
\left({1+\sqrt{1-l_n^2/4}\over2}\right)^{-1}{n\over2}l_n=
{2\over 1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n.
$$
