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I'm trying to prove the transitivty of parallel lines in 3D geometry. ( $l \parallel m \ \land m \parallel n \Rightarrow l \parallel n$ ).

The proof when $l,m,n$ are coplanar is trivial, but I can't seem to come up with a proof when the three lines are non-coplanar.

So, is it possible to prove this fact without the use of any equations or analytic geometry?

PresidentPlant
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  • @AndrewChin I disagree with the use of that duplicate target. At least the accepted answer uses logic that applies in 2D only where as here the OP emphasized that the case of the three lines not being coplanar is of interest. – Jyrki Lahtonen Aug 28 '22 at 08:23

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This is Euclid's Elements Book XI Proposition 9. "(Straight-lines) parallel to the same straight-line, and which are not in the same plane as it, are also parallel to one another."

Pick any point on $m$, and in the two planes $lm$ and $mn$, drop perpendiculars from it onto the lines $l$ and $n$. The point/perpendiculars define a plane to which $m$ is perpendicular (by Proposition 4). And by Proposition 8, if two straight lines are parallel and one is perpendicular to the plane, then so is the other. Hence $l$ and $n$ are perpendicular to the same plane. Finally, Proposition 6 says that if two straight lines ($l$, $n$) are perpendicular to the same plane, then they are parallel.

Reproducing the proofs of Propositions 4, 6, and 8, and everything needed to support them would require a very lengthy answer, and Euclid is a standard work and widely available, so I hope I'll be forgiven for not making this answer self-contained.

Nullius in Verba
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