$\underline{\text{Introduction}:-}$
I think the eyeball theorem is not really that popular and it also doesn't have a Wikipedia page. I have first found it on Quora, however I don't currently remember what the question was, so I cannot link it.
I have also found a question in MO
Searching the name on google also gives some proofs, but those proofs didn't really satisfied me. So I tried proving it on my own.
$\underline{\text{Eyeball Theorem}:-}$
$■\text{Statement}:$
$|GH|=|EF|$
$■\text{My Proof}:-$
(I couldn't find a similar picture, so I just drew it quickly with a sketch pen. If you have problem understanding this picture, then please ask me)
From my picture,
$\angle AI_2P=\angle BI_1R=\angle ACB=\angle ADB=90°$
So,
$\begin{cases}\sin{\alpha}=\frac{RI_1}{BR}=\frac{AC}{AB}\\\sin{\theta}=\frac{PI_2}{AP}=\frac{BD}{AB}\end{cases}$
$\implies\begin{cases}\frac{RI_1}{r_1}=\frac{r_2}{AB}\\\frac{PI_2}{r_2}=\frac{r_1}{AB}\end{cases}$
$\implies\begin{cases}AB\cdot RI_1=r_1r_2\\AB\cdot PI_2=r_1r_2\end{cases}$
$\implies AB\cdot RI_1=AB\cdot PI_2$
$\implies 2RI_2=2PI_2$
$\implies PQ=RS$. $\blacksquare$
$《\bigstar》\text{Generalization}:-$
I think that the $2D$ case implies the $N-D$ case.($N\geq 2$)
Since, the $N$-cones from the centres of the two $N$-balls will create an $N-1$-ball in the places of intersection and since their radii are the same then their $N-1$ volume would be same too which is our concern here.
Correct me if I'm wrong.
$\underline{\text{My Question}:-}$
I would like to have my proof reviewed and I would also like to have more elegant proofs of this theorem.
Best Answer
Your proof seems to be absolutely correct.
But, since you asked for elegancy, I think, it's nicer to not use trignometry in a geometry problem, especially when you can avoid it. Note that the $\sin \theta$ and the $\sin \alpha$ expressions that you invoked were mainly used to find $$\frac{RI_1}{r_1}=\frac{r_2}{AB}\\ \frac{PI_2}{r_2}=\frac{r_1}{AB}$$ which can be very easily and more elegantly derived from noticing the fact that $$\Delta I_2AP\sim \Delta DAB\\ \Delta I_1BR\sim \Delta CBA$$
Also, here's one more proof-
$\underline{\text{My Approach}:-}$
Let $CB$ and $DA$ intersect at $O$ and let $AQ$ and $BS$ extended intersect at $O^\prime$. Notice that $$\Delta COA\sim \Delta DOB$$ Also, because of symmetry, $$AB\perp OO^\prime$$ Using this, we can see that $$\Delta APQ\sim \Delta AOO^\prime\\ \Delta BRS\sim \Delta BOO^\prime$$ The rest is trivial and only requires some algebra with the sides of the similar triangles.