Planes Perpendicular to same Straight Line are Parallel
Theorem
In the words of Euclid:
- Planes to which the same straight line is at right angles will be parallel.
(The Elements: Book $\text{XI}$: Proposition $14$)
Proof
Let $AB$ be a straight line which is perpendicular to each of the planes $CD$ and $EF$.
It is to be demonstrated that $CD$ and $EF$ are parallel.
Suppose, to the contrary, that $CD$ and $EF$ are not parallel.
Then when produced they will meet.
From Proposition $3$ of Book $\text{XI} $: Common Section of Two Planes is Straight Line:
- let the common section be the straight line $GH$.
Let $K$ be an arbitrary point on $GH$.
Let $AK$ and $BK$ be joined.
We have that $AB$ is perpendicular to $EF$.
So from Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:
- $AB$ is perpendicular to the straight line $BK$ in the planes $EF$ produced.
Therefore $\angle ABK$ is a right angle.
For the same reason $\angle BAK$ is a right angle.
Thus, in $\triangle ABK$, there are two angles which are right angles.
From Proposition $17$ of Book $\text{I} $: Two Angles of Triangle are Less than Two Right Angles this is impossible.
Therefore $CD$ and $EF$ do not meet when produced.
So from Book $\text{XI}$ Definition $8$: Parallel Planes:
- $CD$ and $EF$ are parallel.
$\blacksquare$
Historical Note
This proof is Proposition $14$ 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