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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.

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    Lookup [Euclid's fifth postulate](https://en.wikipedia.org/wiki/Parallel_postulate). – 2017-01-08

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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 $\angle QRT=\angle SRU=\pi$

$\ \Rightarrow \angle QRS + \angle SRT= \angle SRT+\angle TRU$

$\ \Rightarrow \angle QRS=\angle TRU=\alpha$

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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 $\angle QPR=\angle PRS=\frac{\pi}{2}$.

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    Indeed, I believe Khan was rather trying to say that the parallel postulate is an axiom that can't be proven from the others. However, this result can be proven from the axioms (but the parallel postulate is needed)... Thanks. – 2017-01-08