If some straight line is drawn parallel to one of the sides of a triangle then it will cut the (other) sides of the triangle proportionally. And if (two of) the sides of a triangle are cut proportionally then the straight line joining the cutting (points) will be parallel to the remaining side of the triangle. * For let $DE$ have been drawn parallel to one of the sides $BC$ of triangle $ABC$. * I say that as $BD$ is to $DA$, so $CE$ (is) to $EA$.
Two triangles ($\bigtriangleup{ABC}$,$\bigtriangleup{ADE}$) are similar if and only if: * they have one congruent angle ($\angle{CAB}\cong\angle{EAD}$) * and two corresponding sides are proportional: $$\frac{|\overline{AB}|}{|\overline{AD}|}=\frac{|\overline{AC}|}{|\overline{AE}|}.$$ The same holds also if this proportion is replaced by one of the following proportionalities: $$\frac{|\overline{AD}|}{|\overline{AE}|}=\frac{|\overline{AB}|}{|\overline{AC}|},\quad\frac{|\overline{AD}|}{|\overline{DB}|}=\frac{|\overline{AE}|}{|\overline{EC}|},\quad\frac{|\overline{AB}|}{|\overline{BC}|}=\frac{|\overline{AD}|}{|\overline{DE}|}.$$
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