in https://math.stackexchange.com/q/1997558/88985 ,
Székely Endre asked too many questions one of the questions was about Bolyai's statement in paragraph 9 of "the Science of absolute space "
in Halsteads translation it goes:
: if $ BN||AM $ and $MAP \bot MAB $ and the $ \angle$ which $NBD$ makes with $NBA$ (on the side of $MABN$ where $MAP$ is) $ < \text{rt} \angle $ then $MAP$ and $NBD$ intersect.
- if (lines)$ BN||AM $
- and (planes) $MAP \bot MAB $
- and the $ \angle$ which (plane) $NBD$ makes with (plane) $NBA$ (on the side of $MABN$ where $MAP$ is) $ < \text{rt} \angle $
Then (the planes) $MAP$ and $NBD$ intersect.
To me it seems to go against the spirit of hyperbolic geometry I always thought than if the distance between the planes $MAP$ and $NBD$ measured on (plane) $MAP$ then the plane will not intersect.
(later) I think I begin to grasp it it is just many different planes are mentioned ( $MAP$, $MAB$ , $NBD$ and $NBA$ )
(more later) there is one plane that contains the points $ A, B, M \text{ and } N $ so the planes $MAB$ and $NBA$ are the same plane) so rewriting the statement it becomes:
- if (lines)$ BN||AM $
- and (planes) $MAP \bot MAB $
- and the $ \angle$ which (plane) $NBD$ makes with (plane) $MAB$ (on the side of $N$ where $MAP$ is) $ < \text{ right } \angle $
Then (the planes) $MAP$ and $NBD$ intersect.
But still can somebody explain (and illustrate) the proof?
Consider two parallel lines $AM$ and $BN$ in the hyperbolic space. Let $AMBN$ denote their common plane as shown below:
Assume that the plane $AMP$ is perpedicular to the plane $AMBN$ and that the dihedral angle, $\delta$ of $BNQ$ and $AMBN$ is between $0$ and $\frac{\pi}2$. ($0<\delta <\frac{\pi}2$).
Bolyai claims, in § 9 of the Appendix, that $AMP$ and $BNQ$ intersect.
His proof goes like this:
Without the loss of generality assume that $AB$ is perpendicular to $AM$. From $A$, drop a perpendicular to $BN$. The intersection point of this perpendicular and $BN$ is denoted by $C$ in the figure above. At $C$ erect a perpendicular $c$ to $BN$ in the plane $BNQ$. The angle of $AC$ and $c$ equals $\delta$.
Now, drop a perpendicular from $A$ to $c$. The right triangle $ACF$ is shown in red above. Obviously $AF<AC$.
Consider now the triangle $ABF$ below:
Here $d$ denotes the intersection line of $ABF$ and $APM$. Now, around $AB$, turn the plane $ABF$ so that $b$ lies in the $AMBN$ plane. Since $AF$<$AC$ the angle $ABF'$ is less than the angle $ABN$. Since $AM$ and $BN$ are parallel $b'$ will meet $AM$. That is, $b$ and $d$ had to meet originally. This proves that the two planes in question have common points -- they intersect.
My note
It seems to be obvious that the angle of $BN$ and $b$ goes to zero as $\delta$ approaches $\frac{\pi}2$. That is, it is not necessary that the intersection points of $AMP$ and $BNQ$ can be found "above" $A$ and $B$. (Above means that "on a perpendicular line hitting $ABMN$ at $A$ or $B$). Considering this remark the statement of § 9. will seem to be more hyperbolic. Take a look at the Klein sphere model of the hyperbolic space:
Here the red plane is perpendicular to the black plane and the blue plane's angle to the black plane is $\delta$. Now, imagine what happens if $\delta$ goes to $\frac{\pi}2$. Clearly the intersection line (purple) of the red and the blue half planes turns upward and finally disappears over the horizon. Very hyperbolic!