Transitivity of parallelism
Dependencies: Through a point off a line, exactly one parallel exists (Playfair).
Statement
Let , , be three lines in the plane. If
then .
Here we adopt the convention "coincident or non-intersecting = parallel" — every line is parallel to itself.

Proof
By contradiction. Suppose . By the definition of parallelism, this means and the two must intersect; let be the intersection point.
Consider the lines through parallel to . On the one hand, since and , is such a line; on the other hand, (parallelism is symmetric) and , so is also such a line.
Hence at least two distinct lines through , namely and , are both parallel to — contradicting uniqueness of parallels. So the original assumption fails, and .

Immediate consequences
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Parallelism is an equivalence relation: reflexivity follows from the definition (every line coincides with itself, hence is parallel to itself); symmetry is a direct consequence of the definition; transitivity is precisely this theorem. Classifying all lines in the plane by "parallel to" partitions them into classes, each called a direction, and "direction" becomes a well-defined notion.
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Direction is preserved under translation: translating the entire figure by some vector sends each line to a line parallel to the original (this can be obtained by the later "draw a parallel through a point"). Combined with this theorem, two parallel lines remain parallel under translation — the foundation for the rigid relations of "parallelograms" later on.
Remarks
The proof uses the symmetry of parallelism (), which follows directly from the symmetric definition "coincident or non-intersecting" — no separate theorem required.
Historically, the transitivity of parallelism was often listed as a direct corollary of the "parallel postulate"; in our system, the substantive work is handed to uniqueness of parallels, and this theorem only uses contradiction to translate "two parallels" into "two parallels through ", at which point uniqueness suffices.