In any triangle, the side opposite to the larger angle is longer

Question

I want to prove the following known theorem

(2) In $\triangle ABC$, if $\angle B>\angle C$, then $AC>AB$

I saw some proofs and they used indirect proof (by contradiction) using the following theorem which seems easier

(1) In $\triangle ABC$, if $AC\leq AB$, then $\angle B\leq\angle C$

Also I saw some proofs using triangle inequality: "sum of each two sides is larger than the other one" which again uses (1).

However I like to prove (2) directly with more elementary methods without using (1) or Pythagorean theorem. Is it possible to do this?

Any comments appreciated.

Answer 1

You certainly do not need the Pythagorean theorem. A "direct proof" will depend on what you have previously proved.

This is Euclid, Book I Proposition 19. The proof there is indirect, depending on the previous proposition. That's the proof you dislike.

Answer 2

Here, $AC=AP+PC=A'P+PC>A'C=AB$.