# 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