Plane through Straight Line Perpendicular to other Plane is Perpendicular to that Plane

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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

Euclid-XI-18.png

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