Planes through Parallel Pairs of Meeting Lines are Parallel

Theorem

In the words of Euclid:

If two straight lines meeting one another be parallel to two straight lines meeting one another, not being in the same plane, the planes through them are parallel.

(The Elements: Book $\text{XI}$: Proposition $15$)

Proof

Let $AB$ and $BC$ be two straight lines which meet each other at $B$.

Let $DE$ and $EF$ be two straight lines which meet each other at $E$ such that:

$AB$ is parallel to $DE$

$BC$ is parallel to $EF$.

It is to be demonstrated that the plane holding $AB$ and $BC$ is parallel to the plane holding $DE$ and $EF$.

From Proposition $11$ of Book $\text{XI} $: Construction of Straight Line Perpendicular to Plane from point not on Plane:

let $BG$ be drawn from $B$ perpendicular to the plane holding $DE$ and $EF$.

From Proposition $31$ of Book $\text{I} $: Construction of Parallel Line:

let $GH$ be drawn from $G$ parallel to $ED$

and

let $GK$ be drawn from $G$ parallel to $EF$.

We have that $BG$ is perpendicular to the plane holding $DE$ and $EF$.

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

$BG$ is perpendicular to all the straight lines which meet it and are in the plane holding $DE$ and $EF$.

But each of $GH$ and $GK$ meets $BG$ and is in the plane holding $DE$ and $EF$.

Therefore each of $\angle BGH$ and $\angle BGK$ is a right angle.

From Proposition $9$ of Book $\text{XI} $: Lines Parallel to Same Line not in Same Plane are Parallel to each other:

$BA$ is parallel to $GH$.

Therefore from Proposition $29$ of Book $\text{I} $: Parallelism implies Supplementary Interior Angles:

$\angle GBA$ and $\angle BGH$ equal two right angles.

But $\angle BGH$ is a right angle.

Therefore $\angle GBA$ is a right angle.

Therefore $GB$ is perpendicular to $BA$.

For the same reason, $GB$ is also perpendicular to $BC$.

We have that the straight line $GB$ is set up perpendicular to the two straight lines $BA$ and $BC$ which cut one another.

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

$GB$ is perpendicular to the plane holding $AB$ and $BC$.

But from Proposition $14$ of Book $\text{XI} $: Planes Perpendicular to same Straight Line are Parallel:

the plane holding $AB$ and $BC$ is parallel to the plane holding $DE$ and $EF$.

$\blacksquare$

Historical Note

This proof is Proposition $15$ 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