I can answer how your question about how this is proved. It assumes the counterclockwise direction and uses vector calculus.
What the proof means by positive angles $is$ that it is counterclockwise direction. In this diagram:
$\theta$ is positive only if you are measuring from $p_1$ to $p_2$, the counterclockwise direction.
As for the proof: there may be other ways, but the way I know uses Green's Theorem:
$$\iint_D \frac{\,dQ}{\,dx}-\frac{\,dP}{\,dy}\,dA=\int_C F\cdot\,dC$$
where $C$ is a parametrized curve in 2 dimensions and $F$ is a vector field that can be represented by $F=(P, Q)$. If $\frac{\,dQ}{\,dx}-\frac{\,dP}{\,dy}=1$, the left hand side of this equation will give you the area. However, as it is impossible to find bounds for this double integral (because there are up to $\infty$ points in the polygon), we use Green's Theorem.
We will use the vector field $(-y, x)$; you will see why in a moment. In this case, $\frac{\,dQ}{\,dx}-\frac{\,dP}{\,dy}=1+1=2$. This gives the equation:
$$\iint_D 2\,dA=\int_C F\cdot\,dC$$
Pulling the 2 out of the double integral and dividing gives $$\iint_D 1 \,dA=A=\frac{1}{2}\int_C F\cdot \,dC$$
Hence the $\frac{1}{2}$ in the Shoelace Formula.
Now the goal is to parametrize the polygon. The line integral previously stated can be broken up into $n$ line integrals along each of the lines connecting two points. In other words:
$$\frac{1}{2}\int_C F\cdot \,dC=\frac{1}{2}\bigg[\sum_{i=1}^{n} \int_{C_i} F\cdot\,dC\bigg]$$
If we can show that one of these line integrals gives ${b_{2}a_1-b_{1}a_2}$, we will have proved the shoelace formula: the $\frac{1}{2}$ distributes out over the sum of the $n$ line integrals.
We will say that $p_1$ is $(a_1, b_1)$ and $p_2$ is $(a_2, b_2)$. A parametrization of this line can be $(a_1, b_1)+t(a_2-a_1, b_2-b_1)$, or
$$x=ta_2-ta_1+a_1\\y=tb_2-tb_1+b_1$$
For $t\in[{0, 1}]$. Good. Now we have our parametrization. If we plug into $F=(-y, x)$, we get $F(t)=(tb_1-tb_2-b_1, ta_2+a_1-ta_1)$.
We also need $\,dC$ in terms of $\,dt$ so we can integrate in respect to $t$. If we take the derivatives of our parametrization in respect to t we get:
$$\frac{\,dx}{\,dt}=\frac{\,d}{\,dt}\big(ta_2-ta_1+a_1\big)=(a_2-a_1)\,dt\\
\frac{\,dy}{\,dt}=\frac{\,d}{\,dt}\big(tb_2-tb_1+b_1\big)=(b_2-b_1)\,dt$$
The dot product is a little tedious, but you get the following:
$$F\cdot\,dC_1=(tb_1-tb_2-b_1, ta_2+a_1-ta_1)\cdot(a_2-a_1,b_2-b_1)\,dt\\=(ta_2b_1-ta_1b_1-ta_2b_2+ta_1b_2-a_2b_1+a_1b_1\\+ta_2b_2-ta_2b_1+a_1b_2-a_1b_1-ta_1b_2+ta_1b_1)\,dt\\
=(a_1b_2-a_2b_1)\,dt$$
At this point you might be able to see where the formula comes from.
Taking the line integral:
$$\int_0^1 (a_1b_2-a_2b_1)\,dt=a_1b_2-a_2b_1.$$
Finally, plugging back into the previous equation gives the shoelace formula:
$$\frac{1}{2}\bigg[\sum_{i=1}^{n}a_ib_{i+1}-a_{i+1}b_i\bigg].\tag*{$\blacksquare$}$$
From the lower left-hand corner of the rectangle (let's call that point $A$), consider the diagonal line as a secant line of the left-hand circle,
intersecting the circle at points $B$ and $C$, where $B$ is between $A$ and $C$.
Also consider one of the edges of the rectangle adjacent to $A$
as a tangent line touching the circle at $D$.
Then by a theorem about tangent and secant lines from a point outside a circle, we have the following relationship of the lengths of the segments
from $A$ to each of the three points $B$, $C$, and $D$:
$$
(AD)^2 = AB \times AC. \tag1
$$
It is easy to find that $AD = \frac12a$.
Now let $E$ be the midpoint of the bottom side of the rectangle;
then $AC$ is the hypotenuse of right triangle $\triangle AEC$,
which has legs $a$ and $\frac12a$, and therefore
$AC = (\frac12\sqrt5)a$.
We can then use Equation $(1)$ to find the length $AB$,
so we can find the length of the chord $BC$; from that chord and
the radius of the circle we can get the angle of $S_1$ at the
center of the circle.
Best Answer
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 $F_{2n}=(1-l_{2n}^2/4)^{-1}f_{2n}$: $$ F_{2n}=\left(1-{l_{2n}^2\over4}\right)^{-1}f_{2n}= \left(1-{2\big(1-\sqrt{1-l_n^2/4}\big)\over4}\right)^{-1}{n\over2}l_n= \left({1\over2}+{1\over2}\sqrt{1-l_n^2/4}\right)^{-1}{n\over2}l_n= \left({1+\sqrt{1-l_n^2/4}\over2}\right)^{-1}{n\over2}l_n= {2\over 1+\sqrt{1-l_n^2/4}}\cdot{n\over2}l_n. $$