Two Lines Perpendicular to Same Plane are Parallel
Theorem
In the words of Euclid:
- If two straight lines be at right angles to the same plane, the straight lines will be parallel.
(The Elements: Book $\text{XI}$: Proposition $6$)
Proof
Let $AB$ and $CD$ be two straight lines at right angles to the plane of reference.
It is to be demonstrated that $AB$ is parallel to $CD$.
Let $AB$ and $CD$ meet the plane of reference at $B$ and $D$ respectively.
Let the straight line $BD$ be joined.
Let $DE$ be drawn in the plane of reference at right angles to $BD$ such that $DE = AB$.
Let $BE$, $AE$ and $AD$ be joined.
We have that $AB$ is at right angles to the plane of reference.
So by Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:
- $AB$ is at right angles to every straight line which meets it and is in the plane of reference.
But each of $BD$ and $BE$ is in the plane of reference and meets $AB$.
Therefore $\angle ABD$ and $\angle ABE$ are both right angles.
For the same reason $\angle CDB$ and $\angle CDE$ are both right angles.
We have that $AB = DE$, and that $BD$ is common.
Thus the two sides $AB$ and $BD$ of $\triangle ABD$ equal the two sides $BD$ and $BE$ of $\triangle EDB$.
The triangles $\triangle ABD$ and $\triangle EDB$ both include right angles.
Therefore by Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:
- $AD = BE$
We have that:
- $AB = DE$
and:
- $AD = BE$
Thus the two sides $AB$ and $BE$ of $\triangle ABE$ equal the two sides $ED$ and $DA$ of $\triangle EDA$.
We also have that $AE$ is common.
So from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:
- $\angle ABE = \angle EDA$
But $\angle ABE$ is a right angle.
Therefore $ED$ is at right angles to $DA$.
But $ED$ is also at right angles to the straight lines $BD$ and $DC$.
Therefore $ED$ is set up at right angles to the three straight lines $BD$, $DA$ and $DC$ at their meeting points.
Therefore from Proposition $5$ of Book $\text{XI} $: Three Intersecting Lines Perpendicular to Another Line are in One Plane:
- $BD$, $DA$ and $DC$ are in the same plane.
But from Proposition $2$ of Book $\text{XI} $: Two Intersecting Straight Lines are in One Plane:
Therefore $AB$ is in the same plane as $DB$ and $DA$.
Therefore the straight lines $AB$, $BD$ and $DC$ are in one plane.
Also, each of $\angle ABD$ and $\angle BDC$ is a right angle.
Therefore from Supplementary Interior Angles implies Parallel Lines:
- $AB$ is parallel to $CD$.
$\blacksquare$
Historical Note
This proof is Proposition $6$ of Book $\text{XI}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions