If two triangles have one angle equal to one angle, and the sides about the equal angles proportional, then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal. * Let $ABC$ and $DEF$ be two triangles having one angle, $BAC$, equal to one angle, $EDF$ (respectively), and the sides about the equal angles proportional, (so that) as $BA$ (is) to $AC$, so $ED$ (is) to $DF$. * I say that triangle $ABC$ is equiangular to triangle $DEF$, and will have angle $ABC$ equal to $DEF$, and (angle) $ACB$ to $DFE$.
Two triangles ($\bigtriangleup{ABC}$,$\bigtriangleup{DEF}$) are similar if: * they have one congruent angle ($\angle{BAC}\cong\angle{EDF}$) * and two corresponding sides are proportional: $$\frac{|\overline{AB}|}{|\overline{AC}|}=\frac{|\overline{DE}|}{|\overline{DF}|}.$$
Proofs: 1