If two triangles have one angle equal to one angle, and the sides about other angles proportional, and the remaining angles either both less than, or both not less than, right angles, then the triangles will be equiangular, and will have the angles about which the sides are proportional equal. * Let $ABC$ and $DEF$ be two triangles having one angle, $BAC$, equal to one angle, $EDF$ (respectively), and the sides about (some) other angles, $ABC$ and $DEF$ (respectively), proportional, (so that) as $AB$ (is) to $BC$, so $DE$ (is) to $EF$, and the remaining (angles) at $C$ and $F$, both less (or not less) than right angles. * I say that triangle $ABC$ is equiangular to triangle $DEF$, and (that) angle $ABC$ will be equal to $DEF$, and (that) the remaining (angle) at $C$ (will be) manifestly equal to the remaining (angle) at $F$.
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: $$\frac{|\overline{AB}|}{|\overline{BC}|}=\frac{|\overline{DE}|}{|\overline{EF}|}.$$
Proofs: 1
