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

Proof

 * Euclid-XI-10.png

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 :
 * $AD = BE$

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 $BE$.

But from :
 * $AD = CF$

and
 * $AD \parallel CF$

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

Therefore from :
 * $AC = DF$

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 :
 * $\angle ABC = \angle DEF$