Three Intersecting Lines Perpendicular to Another Line are in One Plane

Theorem

In the words of Euclid:

If a straight line be set up at right angles to three straight lines which meet one another, at their common point of section, the three straight lines are in one plane.

(The Elements: Book $\text{XI}$: Proposition $5$)

Proof

Let $AB$ be a straight line set up at right angles to the three straight lines $BC$, $BD$ and $BE$ at their point of meeting $B$.

It is to be demonstrated that $BC$, $BD$ and $BE$ are all in the same plane.

Suppose they are not all in the same plane, but that while $BD$ and $BE$ are in the plane of reference, $BC$ is in a plane more elevated.

Thus the plane through $AB$ and $BC$ is different from the plane of reference.

From Proposition $3$ of Book $\text{XI} $: Common Section of Two Planes is Straight Line:

the common section of the plane through $AB$ and $BC$ with the plane of reference is a straight line $BF$, say.

Therefore the three straight lines $AB$, $BC$ and $BF$ are in the same plane, that is, the one through $AB$ and $BC$.

We have that $AB$ is at right angles to each of the straight lines $BD$ and $BE$.

Therefore from Proposition $4$ of Book $\text{XI} $: Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane:

$AB$ is at right angles to the plane through $BD$ and $BE$.

But the plane through $BD$ and $BE$ is the plane of reference.

Therefore $AB$ is at right angles to the plane of reference.

Thus from Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:

$AB$ will also make right angles with all the straight lines which meet it and are in the plane of reference.

But $BF$, which is in the plane of reference, meets $AB$.

Therefore $\angle ABF$ is a right angle.

But by hypothesis $\angle ABC$ is also a right angle.

Therefore $\angle ABF = \angle ABC$.

But $\angle ABF$ and $\angle ABC$ are in one plane, which is impossible.

Therefore $BC$ is not in a plane more elevated.

Therefore $BC$, $BD$ and $BE$ are all in the same plane.

$\blacksquare$

Historical Note

This proof is Proposition $5$ 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