At 4:43 of the following video Khan from Khan Academy states that there is no proof that two parallel lines crossed by a transversal line have the same angle. Rather it's to be taken on "intuition".
https://www.youtube.com/watch?v=H-E5rlpCVu4&list=PL26812DF9846578C3&index=12
That is interesting - surely there must be a proof for this rather obvious result, or otherwise it must be an axiom taken on faith? I haven't seen euclidean geometry having such an axiom. To make matters worse, it seems that almost all interesting results of geometry are dependent on this result.
Picture of the problem:
The task is to demonstrate that the angles marked by alpha are indeed the same, no matter what the bottom angle is.
Drop perpendicular from $R$ on line $cd$.
$\triangle PQR$ is right angled $\Rightarrow \angle PRQ= \frac{\pi}{2}-\alpha$.
Since $\angle PRS=\frac{\pi}{2} \Rightarrow \angle QRS=\alpha$.
Also $\angle QRT=\angle SRU=\pi$
$\ \Rightarrow \angle QRS + \angle SRT= \angle SRT+\angle TRU$
$\ \Rightarrow \angle QRS=\angle TRU=\alpha$
Note:
Incase you are wondering that how can I say that $\angle QPR=\angle PRS=\frac{\pi}{2}$.
If $\angle QPR\ne\angle PRS=\frac{\pi}{2}$ then the lines $ab$ and $cd$ wont be parallel and would intersect at some point.
Hence $\angle QPR=\angle PRS=\frac{\pi}{2}$.