I am trying to understand the short proof of Fuss' Formula in the paper "Bicentric Quadrilaterals through Inversion" by Albrecht Hess which is available here:
https://forumgeom.fau.edu/FG2013volume13/FG201303.pdf
The proof is given in section 3, but there are two points in the second paragraph that I don't understand. First, I don't see why $B'SD'Z$ is a parallelogram as claimed and also I dont see why $2ZD'^2 + 2SD'^2 = 2r^2$.
I would appreciate if someone could elaborate on these points.
Let us first prove that $2ZD'^2 + 2SD'^2 = 2r^2$.
$EG$ and $FH$ meet at $S$. (A proof is written at the end of this answer.)
$EG$ is parallel to $A'D'$, and $FH$ is parallel to $C'D'$. Since $A'B'C'D'$ is a rectangle, we see that $\angle{GSH}=90^\circ$. Since $D'$ is the midpoint of $GH$, we see that $D'$ is the center of a circle on which $G,H,S$ exist. So, it follows that $SD'=D'G$.
Therefore, using the fact that $\triangle{ZD'G}$ is a right triangle, we finally get $$2(ZD'^2+SD'^2)=2(ZD'^2+D'G^2)=2\times ZG^2=2r^2\tag1$$
Next, let us prove that $B'SD'Z$ is a parallelogram.
Let us consider $\triangle{HD'S}$. Since $\triangle{HD'S}$ is an isosceles triangle with $\angle{ZD'H}=90^\circ$, we get $$\angle{ZD'S}=90^\circ-2\angle{D'HS}\tag2$$ Similarly, considering $\triangle{SB'E}$, we get $$\angle{ZB'S}=90^\circ-2\angle{SEB'}\tag3$$ We also have $$\angle{D'HS}=\angle{SEB'}\tag4$$ because $\angle{D'HS}=\angle{GHF}=\angle{GEF}=\angle{SEB'}$. It follows from $(2)(3)(4)$ that $$\angle{ZD'S}=\angle{ZB'S}\tag5$$
Let $D'Z=a,B'Z=b$ and $SZ=c$. Then, it follows from $(1)$ that $D'S=\sqrt{r^2-a^2}$. Similarly, $B'S=\sqrt{r^2-b^2}$. Applying the law of cosines to $\triangle{SD'Z}$, we get $$\cos\angle{ZD'S}=\frac{r^2-c^2}{2a\sqrt{r^2-a^2}}\tag6$$ Similarly, applying the law of cosines to $\triangle{ZB'S}$, we get $$\cos\angle{ZB'S}=\frac{r^2-c^2}{2b\sqrt{r^2-b^2}}\tag7$$
It follows from $(5)(6)(7)$ that $\frac{r^2-c^2}{2a\sqrt{r^2-a^2}}=\frac{r^2-c^2}{2b\sqrt{r^2-b^2}}$. Squaring the both sides of $a\sqrt{r^2-a^2}=b\sqrt{r^2-b^2}$, we get $r^2=a^2+b^2$.
So, $D'S=B'Z$ and $B'S=D'Z$. Since $\triangle{D'ZS}\equiv\triangle{B'SZ}$, we get $\angle{D'SZ}=\angle{B'ZS}$. It follows that $D'S$ is parallel to $B'Z$. So, we can say that $B'SD'Z$ is a parallelogram.
Finally, let us prove that $EG$ and $FH$ meet at $S$.
Proof :
Let $O'$ be the intersection point of $HF$ and $GE$. Let us first show that $BD$ passes through $O'$.
Let $J$ be a point on $GE$ such that $BJ$ is parallel to $CD$. Also, let $K$ be a point on $HF$ such that $BK$ is parallel to $AD$.
As shown below, we see that $\triangle{BJK}$ is an isosceles triangle, and that $\triangle{DGH}\sim\triangle{BJK}$.
Since $O'$ is a homothetic center for the two triangles $\triangle{DGH}$ and $\triangle{BJK}$, $BD$ passes through $O'$. Similarly, $AC$ passes through $O'$.
Therefore, $EG$ and $FH$ meet at $S$.$\ \blacksquare$