Proof for similarities between two triangles.

Question

We know that if the angles of two triangles are similar, then their sides are proportional. I get the idea. Now, can it be proven rigorously?

Answer 1

Through a rigid motion, you can have angles $\widehat{BAC}$ and $\widehat{B'A'C'}$ coincide with additional conditions $A'=A$, $B'\in l(AB)$ and $C'\in l(AC)$. By using the converse of the parallel lines theorem on $l(B'C')$ and $l(BC)$ with transversal $l(AB)$, we deduce that $B'C'\parallel BC$. Consider now $r\parallel l(BC)$ such that $A=A'\in r$. By Thales' theorem on three parallel lines and two transverse lines, $$A'B':AB=A'C':AC$$

Now that you know this proportionality, the two triangles $\triangle ABC$ and $\triangle A'B'C'$ fall into the SAS criterion for proportionality.