(Straight-lines) parallel to the same straight-line are also parallel to one another.  Let each of the (straight-lines) $AB$ and $CD$ be parallel to $EF$. I say that $AB$ is also parallel to $CD$.  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.