# Proving the AIP theorem?#2

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.

• 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"). – vociferous_rutabaga Sep 1 '15 at 23:49
• 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.) – Blue Sep 2 '15 at 1:23
• 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). – Aretino Sep 2 '15 at 19:49

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.