Through a point off a line, exactly one parallel exists (Playfair)

Dependencies: Corresponding/alternate angles ⇔ lines parallel (Corresponding/alternate angles ⇔ lines parallel), dropping a perpendicular from a point to a line (Perpendicular from a point to a line exists and is unique), the triangle exterior-angle inequality (Exterior angle > either non-adjacent interior angle).

Statement

Let $\ell$ be a line, and let $P$ be a point off $\ell$ ($P\notin\ell$). Then there exists exactly one line through $P$ parallel to $\ell$.

Denote this unique parallel by $n$. The statement has two halves: existence — at least one can be constructed; uniqueness — there is no second one. This is the celebrated Playfair's axiom of elementary geometry — equivalent in Euclidean geometry to Euclid's fifth postulate; here we treat it as a theorem to be proved, using only the three already-established results above.

Proof

Existence — two perpendiculars in succession. By Perpendicular from a point to a line exists and is unique, drop a perpendicular from $P$ to $\ell$, with foot $F$, giving the line $m\perp\ell$ at $F$ (and $P\in m$). Apply Perpendicular from a point to a line exists and is unique again at the point $P$ on $m$ to draw the perpendicular to $m$, giving a line $n\perp m$ with $P\in n$. View $m$ as a transversal cutting $\ell$ and $n$: at $F$, $m\perp\ell$ gives a $90^\circ$ angle; at $P$, $m\perp n$ gives a $90^\circ$ angle; the two lie on the same side of $m$ and form a pair of equal corresponding angles. By Corresponding/alternate angles ⇔ lines parallel we obtain $n\parallel\ell$.

Uniqueness — contradiction + Exterior angle > either non-adjacent interior angle. Suppose another line $\ell'\neq n$ also passes through $P$ with $\ell'\parallel\ell$. Examine the angle between $\ell'$ and $m$ at $P$: if this angle $=90^\circ$, then $\ell'\perp m$ at $P$; but $n$ is already the perpendicular to $m$ at $P$, so by the uniqueness part of Perpendicular from a point to a line exists and is unique we get $\ell'=n$, contradicting the assumption. Hence this angle $\neq 90^\circ$. Without loss of generality, let $\theta$ denote the acute angle (so $\theta<90^\circ$) among the two pairs of vertical angles formed by $\ell'$ and $m$ at $P$, and let $r$ be the ray on $\ell'$ from $P$ entering "the half-plane containing $\ell$ (i.e. the side containing $F$)" such that the angle between $r$ and the ray $PF$ is exactly $\theta$ (this is the side on which $\theta$ appears; the other side carries the supplement of $\theta$).

Key step — pick a triangle big enough. Starting from the ray $PF$, walk far enough along $\ell$ to a point $F''\in\ell$ so that the interior angle $\angle FPF''$ of $\triangle PFF''$ at $P$ is greater than $\theta$. This step is delivered by the protractor axiom (axiom III): as $F''$ moves along $\ell$ away from $F$, $\angle FPF''$ takes every value in $[0^\circ,90^\circ)$ and approaches $90^\circ$; since $\theta<90^\circ$, there exists $F''$ with $\angle FPF''>\theta$. Fix such an $F''$.

Examine $\triangle PFF''$. At $F$, $\angle PFF''=90^\circ$ (because $m\perp\ell$). At $P$, the angle between rays $PF$ and $PF''$ is $\angle FPF''>\theta$, while the angle between $r$ and ray $PF$ is exactly $\theta$; moreover $r$ and ray $PF''$ lie on the same side of ray $PF$ (both are in the half-plane containing $\ell$, on $F$'s side), so $r$ lies inside $\angle FPF''$ (angle additivity, axiom III). In other words, the ray $r$ starts at $P$, lies along $\ell'$, and enters the interior of $\triangle PFF''$.

By a standard corollary of Exterior angle > either non-adjacent interior angle (also known as the Pasch lemma): a ray from a vertex of a triangle that enters the interior must meet the opposite side — here the "opposite side" is $FF''\subset\ell$. Hence $r$ must meet $\ell$. But $r\subset\ell'$, so $\ell'$ meets $\ell$, directly contradicting $\ell'\parallel\ell$.

In sum, the only line through $P$ parallel to $\ell$ is $n$. $\blacksquare$

Immediate consequences

• Parallel is transitive: if $a\parallel b$ and $b\parallel c$ ($a,b,c$ pairwise distinct), then $a\parallel c$. Otherwise $a\cap c\ni P$, and through $P$ both $a$ and $c$ would be parallel to $b$ — contradicting parallel uniqueness. This transitivity is the foundation of subsequent arguments about parallelograms and similar figures.
• A prototype of the converse corresponding-angles theorem: if two parallel lines are cut by a third line, the corresponding angles are equal. Proof by contradiction using parallel uniqueness: if the corresponding angles differ, at the intersection point one can draw another line whose corresponding angle with the transversal equals that of the first line; by Corresponding/alternate angles ⇔ lines parallel this new line is also parallel to the first, but it passes through the same intersection point parallel to the same line — contradicting parallel uniqueness.
• Triangle angle sum $=180^\circ$: the classical proof draws through a vertex of the triangle a parallel to the opposite side, transporting the other two vertex angles to the vertex via corresponding and alternate interior angles to combine into a straight line. The "unique parallel auxiliary line" used for the transport is selected by parallel uniqueness.

Remarks

Playfair's place in the system — in Euclidean systems this conclusion has two equivalent "legislative roles": either write down Playfair's axiom as a postulate; or, as in this section, prove it from the three combined results Perpendicular from a point to a line exists and is unique (perpendicular existence and uniqueness) + Corresponding/alternate angles ⇔ lines parallel (equal corresponding angles ⇒ parallel) + Exterior angle > either non-adjacent interior angle (which blocks "another parallel"). The latter reveals that the genuine hardness of parallel uniqueness is concentrated in Corresponding/alternate angles ⇔ lines parallel's "equal corresponding angles ⇒ parallel" — once we move to hyperbolic geometry, Exterior angle > either non-adjacent interior angle still holds but Corresponding/alternate angles ⇔ lines parallel fails, and the parallels through $P$ become infinite in number.

Two-stage structure of the uniqueness proof — there are two small details worth peeling off in the proof. First, the boundary case "$\ell'\perp m$" is peeled off and handed to the uniqueness part of Perpendicular from a point to a line exists and is unique; the remaining case "$\ell'$ not perpendicular to $m$" is then docked with Exterior angle > either non-adjacent interior angle. Second, an early draft tried to read off "the exterior angle of $\triangle PFF'$ at $P$ $=\theta$", but this equality does not hold in elementary diagrams — the side $PF'$ does not lie along $\ell'$ ($F'$ is the projection of $Q$ onto $\ell$, not on $\ell'$), so the "exterior angle" of $\triangle PFF'$ and the "angle between $\ell'$ and $m$" are two different angles. We correct this to "choose a sufficiently large $\triangle PFF''$ so that $r$ enters its interior, then apply the ray–opposite-side lemma", which is one of the standard corollaries of Exterior angle > either non-adjacent interior angle (a ray entering the interior of a triangle must meet the opposite side), closing the logical chain entirely.

Comparison with hyperbolic geometry — worth recording: in hyperbolic geometry, infinitely many lines through $P$ are parallel to $\ell$, and every line between the "two limiting parallels" fails to meet $\ell$. The "choose $F''$ with $\angle FPF''>\theta$" step in this section still works in the hyperbolic setting (the protractor axiom is neutral), and the corollary of Exterior angle > either non-adjacent interior angle "$r$ enters $\triangle PFF''$ ⇒ $r$ meets the opposite side" also still holds — what fails in the hyperbolic setting is not this step, but rather the step "the existence of another $\ell'\parallel\ell$ pair can be ruled out directly by Corresponding/alternate angles ⇔ lines parallel": in the hyperbolic setting, an angle no longer uniquely determines a parallel direction.