# Proof of “triangles are similar iff corresponding angles are equal”

I'm looking for a basic proof of the basic plane Euclidean geometry theorem: Two triangles are similar if and only if their corresponding (interior) angles are equal.

(The theorem can be also worded in terms of only two of the interior angles.)

I'm having great difficulty finding an "elementary" proof of this theorem. (I have searched both online and in my local university library.)

By "elementary" I mean a proof that uses only "the standard axioms" of plane Euclidean geometry (as taught in basic high-school geometry). (In particular, I'm specifically ruling out proofs that are based on analytic geometry.)

I have been able to find some proofs online, but they all depend on the theorem that "two transversals are divided into proportional segments by a system of parallel lines." Unfortunately, every proof that I can find of this particular theorem happens to be based on the subject line's theorem.

Can someone point me to a reference where this theorem is proved in full?

EDIT:

When I write "two triangles $ABC$ and $DEF$ are similar" I mean that the ratios of the lengths of their (corresponding) sides are equal, i.e.,

$$\overline{AB}:\overline{BC}:\overline{CA} = \overline{DE}:\overline{EF}:\overline{FD}$$

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What is your definition for two triangles to be similar? If it is not that they have corresponding angles, I take it that is that the triangles have sidelengths in the same ratio? –  JavaMan Jun 16 '13 at 17:09
(@JavaMan: see my EDIT.) –  kjo Jun 16 '13 at 18:04