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How do you prove the AIP theorem?

(Alternate Interior Angles and Parallel theorem) I already know you can prove the CAP theorem(Corresponding Angles and Parallel theorem), and the SAP theorem from it,but I don't know how to prove the AIP as a start.

The AIP Theorem.

Given: If transversal $n$ intersects two lines in such a way that a pair of alternate exterior angles are equal, then the two lines are parallel.

Statement: Line $n$ intersects lines $l$ and $m$ such that a pair of alternate exterior angles are congruent.

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    $\begingroup$ Can you include complete names or definitions instead of acronyms? Some of us aren't familiar with the abbreviations AIP and CAP :) (And a quick google for "AIP" resulted in "American Institute of Physics"). $\endgroup$ – vociferous_rutabaga Sep 1 '15 at 23:49
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    $\begingroup$ You're joking with the "Consistency, Availability, and Partition tolerance" thing, right? I presume that, since AIP is (more-or-less) the Alternate Interior Angles and Parallelism theorem, then CAP should be the Corresponding Angles and Parallelism theorem, regarding angle pairs like $\angle 4$ and $\angle 8$. (Personally, I call the Alternate Interior Angles theorem the "Z" theorem, and the Corresponding Angles theorem the "F" theorem.) $\endgroup$ – Blue Sep 2 '15 at 1:23
  • $\begingroup$ It all depends on the axioms you have chosen. Usually this theorem follows trivially by the external angle theorem (an external angle of a triangle is greater than any of the internal angles not adjacent to it). $\endgroup$ – Aretino Sep 2 '15 at 19:49
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If $\angle 3 = \angle 6$, then by the vertical angles theorem (even seeing the name, I have no clue what the CAP theorem is supposed to be) $\angle 6 = \angle 3 = \angle 2$, whence $\angle 6$ and $\angle 4$ are supplementary, which if $m$ and $n$ intersect on that side would form a triangle with the third angle at the intersection being $0$ degrees, which cannot be. So they cannot intersect on the side of angles 4 and 6. Similarly, they cannot intersect on the side of angles 3 and 5. Therefore they cannot intersect.

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Its like this: First, you assume l is not parallel to m. Then they inersect and form a triangle. By the exterior angle theoreom (a exterior angle is bigger than its remote interior angles), you have one of the alternate interior angle is larger than the other, and therefore are not congruent. This contradicts the given (the alternate interior angle are congruent) and so the l is not parallel to m assumption is false, since it caused the contradiction. Therefore l is parallel to m.

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