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