Plane through Straight Line Perpendicular to other Plane is Perpendicular to that Plane
Theorem
In the words of Euclid:
- If a straight line be at right angles to any plane, all the planes through it will be at right angles to the same plane.
(The Elements: Book $\text{XI}$: Proposition $18$)
Proof
Let $AB$ be an arbitrary straight line which is perpendicular to the plane of reference.
It is to be demonstrated that every plane holding $AB$ is perpendicular to the plane of reference.
Let an arbitrary plane $DE$ be drawn through $AB$.
Let $CE$ be the common section of the plane $DE$ and the plane of reference.
Let $F$ be an arbitrary point on $CE$.
From Proposition $11$ of Book $\text{I} $: Construction of Perpendicular Line:
- let $FG$ be drawn in $DE$ perpendicular to $CE$.
We have that $AB$ is perpendicular to the plane of reference.
Therefore from Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:
- $AB$ is perpendicular to all the straight lines which meet it and are in the plane of reference.
Therefore $AB$ is perpendicular to $CE$.
Therefore $\angle ABF$ is a right angle.
But $\angle GFB$ is also a right angle.
Therefore by Proposition $28$ of Book $\text{I} $: Supplementary Interior Angles implies Parallel Lines:
- $AB \parallel FG$
But $AB$ is perpendicular to the plane of reference.
Therefore from Proposition $8$ of Book $\text{XI} $: Line Parallel to Perpendicular Line to Plane is Perpendicular to Same Plane:
- $FG$ is perpendicular to the plane of reference.
Thus the conditions are fulfilled for Book $\text{XI}$ Definition $4$: Plane at Right Angles to Plane to apply:
- $DE$ is perpendicular to the plane of reference.
As $DE$ is arbitrary, the result follows.
$\blacksquare$
Historical Note
This proof is Proposition $18$ 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