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