Prove that alternate interior and exterior angles are equal if the lines who are crossed by the third line is parallel.
Could someone review my proof? I used a proof by contradiction.
- $\alpha+\delta=180^{\circ}$ (straight angle)
- Say lines $l$ and $m$ intersect in a point $C$.
- Then in $\triangle ABC$ $\angle A+\angle B+\angle C=180^{\circ} \Rightarrow \angle C=0^{\circ}$
- Contradiction, because $\angle C$ cannot be $0^{\circ}$.
I am not sure about the last line. Could someone verify that it is OK?
This is (basically) a valid proof by contradiction that
It would be best if you added "assume the opposite interior angles are equal" into the lines of your proof. I can see you did that in your picture but it would really be best to make the writing independent of the picture. Pictures like this can help scaffold proofs but they should really be disposable at the end.
Nothing at all is said about the only if direction.
Edit: Removed a f in my post. – monkey1234 10 secs ago
Nothing has become untrue in my solution.
Edit: My question was rather: is it valid to say that in a triangle on of the angles cannot be $0^\circ$? – monkey1234 7 secs ago
Yes: in metric geometry, two distinct lines cannot intersect at an angle of $0^\circ$. Your hypotheses no doubt prevent $m$ and $\ell$ from being the same line.