# Two Lines Meeting which are Parallel to Two Other Lines Meeting contain Equal Angles

## Theorem

In the words of Euclid:

*If two straight lines meeting one another be parallel to two straight lines meeting one another not in the same plane, they will contain equal angles.*

(*The Elements*: Book $\text{XI}$: Proposition $10$)

## Proof

Let $AB$ and $BC$ be straight lines which meet one another.

Let $DE$ and $EF$ be straight lines which meet one another such that $AB$ is parallel to $DE$ and $BC$ is parallel to $EF$.

It is to be demonstrated that $\angle ABC = \angle DEF$.

Let $BA, BC, ED, EF$ be cut off equal to one another.

Let $AC, CF, BE, AD, DF$ be joined.

We have that:

- $BA = ED$

and

- $BA \parallel ED$

Therefore from Proposition $33$ of Book $\text{I} $: Lines Joining Equal and Parallel Straight Lines are Parallel:

- $AD = BE$

and

- $AD \parallel BE$

For the same reason:

- $CF = BE$

and

- $CF \parallel BE$

So each of $AD$ and $CF$ is equal and parallel to $BE$.

- $AD = CF$

and

- $AD \parallel CF$

We have that $AC$ and $DF$ join $AD$ and $CF$.

Therefore from Proposition $33$ of Book $\text{I} $: Lines Joining Equal and Parallel Straight Lines are Parallel:

- $AC = DF$

and

- $AC \parallel DF$

So we have that $AB$ and $BC$ are equal and parallel to $DE$ and $EF$.

We also have that $AC = DF$.

So from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:

- $\angle ABC = \angle DEF$

$\blacksquare$

## Historical Note

This proof is Proposition $10$ 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