Proof: By Euclid

For let the straight line $GK$ fall across ($AB$, $CD$, and $EF$).
And since the straight line $GK$ has fallen across the parallel straight lines $AB$ and $EF$, (angle) $AGK$ (is) thus equal to $GHF$ [Prop. 1.29].
Again, since the straight line $GK$ has fallen across the parallel straight lines $EF$ and $CD$, (angle) $GHF$ is equal to $GKD$ [Prop. 1.29].
But $AGK$ was also shown (to be) equal to $GHF$.
Thus, $AGK$ is also equal to $GKD$.
And they are alternate (angles).
Thus, $AB$ is parallel to $CD$ [Prop. 1.27].
Thus, (straight lines) parallel to the same straight line are also parallel to one another.
(Which is) the very thing it was required to show.
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