In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. And those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal. * Let $AB$ and $BC$ be equal and equiangular parallelograms having the angles at $B$ equal. * And let $DB$ and $BE$ be laid down straight-on (with respect to one another). * Thus, $FB$ and $BG$ are also straight-on (with respect to one another) [Prop. 1.14]. * I say that the sides of $AB$ and $BC$ about the equal angles are reciprocally proportional, that is to say, that as $DB$ is to $BE$, so $GB$ (is) to $BF$.
Two parallelograms are congruent if and only if their angles and the products^{1} of the side lengths of an angle are in both paralleograms equal.
$$\begin{array}{rclc} \boxdot{ADBF}\cong\boxdot{BCEG}&\Longleftrightarrow&(\angle{ADB}=\angle{BGC})&\wedge\\ &&(\angle{FAD}=\angle{EBG})&\wedge\\ &&(|\overline{DB}|\cdot|\overline{BF}|=|\overline{BE}|\cdot|\overline{GB}|). \end{array}$$
Proofs: 1
The product is equivalent to Euclid's "reciprocal proportion" $$\frac{|\overline{DB}|}{|\overline{BE}|}=\frac{|\overline{GB}|}{|\overline{BF}|},$$ which is not to be confused with the concept of reciprocal ratio. ↩