From Isaacs, Geometry for College Students.
(1.28) Theorem. If Triangle $ABC$ is similar to Triangle $XYZ$ (i.e. the angles match up) then the corresponding side lengths are proportional.
Now as I recall, this fact is established in high school geometry and I expected an easy proof. Certainly it is "obvious" if you think of embedding the two triangles in coordinate space and using trig.
However, this book is avoiding coordinates and trigonometry. Usually that makes for prettier proofs (certainly it does in this book), but I'm not satisfied with Isaacs' proof.
The key is the following.
(1.29) Lemma. Let $U$ and $V$ be points on sides $AB$ and $AC$ of triangle $ABC$. Then $UV$ is parallel to $BC$ if and only if $AU/AB = AV/AC$.
The proof of this lemma boils down to proving that triangles $BUC$ and $BVC$ have equal areas. It's some work to do this, it's a little more work to show that this implies the lemma, and it's still a little bit more work to use this to prove the main theorem.
Altogether the proof is more complicated than I was expecting, and I find it vaguely disquieting that the proof appeals to facts about areas, which seems like conceptual overkill. Nevertheless I am unable to think of a simpler or intuitive proof, at least without appealing to coordinates in some form. Indeed, I was expecting Isaacs to just postulate this, as he does (for example) for the SAS triangle congruence condition.
Is there some context which I haven't considered in which this is the "right" proof?
