In the 3-dimensional hyperbolic space there are given a plane $\mathcal{P}$ and four distinct lines $a\_1, a\_2, r\_1, r\_2$ in such positions that $a\_1$ and $a\_2$ are perpendicular to $\mathcal{P}$, $r\_1$ is coplanar with $a\_1, r\_2$ is coplanar with $a\_2$, finally $r\_1$ and $r\_2$ intersect $\mathcal{P}$ at the same angle. Rotate $r\_1$ around $a\_1$ and rotate $r\_2$ around $a\_2$; denote by $\mathcal{S}\_1$ and $\mathcal{S}\_2$ the two surfaces of revolution they sweep out. Show that the common points of $\mathcal{S}\_1$ and $\mathcal{S}\_2$ lie in a plane.



3-dimensional hyperbolic space
Let $L$ be the geodesic segment in $P$ with one endpoint at $a\_1 \cap P$ and the other endpoint at $a\_2 \cap P$. Let $b$ be the midpoint of $L$. We define a plane $Q$ by requiring it to contain $b$ and by requiring it to be perpendicular to $L$ (and thus to $P$). Thus reflection in $Q$ exchanges $a\_1$ and $a\_2$. It is now an exercise to prove that reflection in $Q$ exchanges the “cones” $S\_1$ and $S\_2$. Thus $S\_1 \cap S\_2$ lies in the plane $Q$.




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Here I am assuming that the lines $r\_1$ and $r\_2$ actually meet $P$ — the question is slightly unclear on this detail.


