Perp. unique — Perpendicular from a point to a line exists and is unique · Principia

Dependencies: Protractor axiom, Linear pair sums to 180°, SAS congruence, Exterior angle > either non-adjacent interior angle.

Statement

Let $\ell$ be a line and $P$ any point in the plane (either $P\in\ell$ or $P\notin\ell$ ). Then there exists and is unique a line $\ell'$ through $P$ that is perpendicular to $\ell$ .

$Unified statement of the two cases: on the left, when P\in\ell, take the 90^\circ ray directly via the protractor to obtain \ell'; on the right, when P\notin\ell, construct \ell' through P perpendicular to \ell at the foot F.$

Proof

Existence · Case 1 ( $P\in\ell$ ). Place the centre of the protractor at $O = P$ with the initial side along one of the rays of $\ell$ . By Protractor axiom, take the ray $r$ at the value $90^\circ$ ; $r$ together with its opposite ray makes a line $\ell'$ . The angle between $\ell'$ and $\ell$ at $P$ is exactly $90^\circ$ , hence $\ell'\perp\ell$ .

Existence · Case 2 ( $P\notin\ell$ ). Pick any $A$ on $\ell$ . By Protractor axiom, on the other side of $\ell$ construct a ray with $\angle QA\ell = \angle PA\ell$ , then by the Ruler axiom pick $Q$ on this ray with $AQ = AP$ . Since $P$ and $Q$ lie on opposite sides of $\ell$ , the segment $PQ$ meets $\ell$ at some point $F$ . In $\triangle PAF$ and $\triangle QAF$ , we have $AP = AQ$ , $\angle PAF = \angle QAF$ , and the common side $AF$ , so by SAS congruence $\triangle PAF\cong\triangle QAF$ , hence $\angle PFA = \angle QFA$ . Also $\angle PFA + \angle QFA = 180^\circ$ (Linear pair sums to 180°), so

\angle PFA = \angle QFA = 90^\circ,

i.e. the line $PQ$ passes through $P$ and is $\perp\ell$ .

$Existence in Case 2: reflect P across \ell to Q; \triangle PAF \cong \triangle QAF (SAS), and combining with the linear-pair sum 180^\circ yields \angle PFA = \angle QFA = 90^\circ.$

Uniqueness. Suppose there is another line $m$ through $P$ with $m\perp\ell$ at some $G\neq F$ . In $\triangle PFG$ , we have $\angle PFG = 90^\circ$ ; viewing $\angle PG\ell$ as the exterior angle of this triangle at $G$ , by Exterior angle > either non-adjacent interior angle it is strictly greater than the non-adjacent interior angle $\angle PFG = 90^\circ$ . But $m\perp\ell$ gives $\angle PG\ell = 90^\circ$ , contradicting $>90^\circ$ , so the perpendicular through $P$ is unique. $\blacksquare$

$Uniqueness by contradiction: suppose a second line m\perp\ell at G\neq F; in \triangle PFG we have \angle PFG = 90^\circ, but the exterior angle at G along \ell must be > 90^\circ (exterior-angle inequality), contradicting the assumed 90^\circ.$

Immediate consequences

Uniqueness of parallels (a key lemma for Through a point off a line, exactly one parallel exists (Playfair)): this theorem makes "drop a perpendicular from $P$ to $\ell$ " explicit, providing the geometric prerequisite for constructing and comparing in subsequent proofs that "at most one line through $P$ is parallel to $\ell$ ".
Perpendicular segment is shortest (Perpendicular segment is shortest): the unique perpendicular makes "the distance from $P$ to $\ell$ $= |PF|$ " well-defined; every argument of the form "the perpendicular segment is shorter than any oblique segment" rests on this theorem.

Remarks

Existence in the two cases shares a common origin — Protractor axiom generates directions by numerical value: in Case 1 simply take $90^\circ$ ; in Case 2 first construct a symmetric point $Q$ on the other side, then let SAS translate "equal angles on both sides + linear-pair sum $180^\circ$ " into "each equals $90^\circ$ ". The uniqueness step goes deeper: it depends on Exterior angle > either non-adjacent interior angle rather than the stronger "interior-angle sum $= 180^\circ$ " — the latter requires the parallel postulate, whereas the present theorem is a neutral-geometry result, valid in geometries on both sides.