If two triangles ABC and A'B'C' have equal corresponding angles, use Thales Theorem to prove that $\frac {|AB|}{|A'B'|}=\frac{|BC|}{|B'C'|}=\frac{|AC|}{|A'C'|}$.
I know that we can place the smaller triangle (say ABC) inside of the larger one with one of the angles lining up. The issue that I'm coming across is showing that the bases of the two triangles are parallel.
Take $K\in B'C'$ such that $B'BCK$ is a parallelogram and use Thales.