平行线的传递性

The theorem is: If line $l$ $||$ $m$ and line $m$ $||$ $n$, then line $l$ $||$ $n$.

Here's the proof: We are given that $l$ $||$ $m$ and line $m$ $||$ $n$. Take a transversal $t$ intersecting all three lines. Now, as you know, with parallel lines, corresponding angles are congruent. So, ding! Here's a pair of corresponding angles between $l$ and $m$. Ding! Another pair, with $m$ and $n$. Now, it is important to make sure that there is an overlapping angle on line $m$ between the two pairs. So guess what? Using the two angles only in one pair, from $l$ and $n$, we can make another pair of corresponding angles! Now don't forget: the corresponding angle postulate is an iff statement: it works both ways. So by seeing that pair of corresponding angles from $l$ and $n$, that means $l$ $||$ $n$, as we wanted to show.