Equiangular parallelograms have to one another the ratio compounded^{1} out of (the ratios of) their sides. * Let $AC$ and $CF$ be equiangular parallelograms having angle $BCD$ equal to $ECG$. * I say that parallelogram $AC$ has to parallelogram $CF$ the ratio compounded out of (the ratios of) their sides.
The proportion of the areas of equiangular parallelograms equals the product of proportions of their sides building the legs of equal angles. More precisely, if $\boxdot{ABCD}$ and $\boxdot{CEFG}$ are parallelograms with equal angles $\gamma:=\angle{BCD}$ (with the legs of lengths $a:|\overline{BC}|,$ $b:=|\overline{DC}|$) and $\gamma':=\angle{BCD}$ (with the legs of lengths $a':|\overline{CG}|,$ $b':=|\overline{CE}|$), then the following formula holds:
$$\frac{\operatorname{area}\boxdot{ABCD}}{\operatorname{area}\boxdot{CEFG}}=\frac{ab\sin(\gamma)}{a'b'\sin(\gamma')}=\frac{a}{a'}\cdot\frac{b}{b'}.$$
Proofs: 1
In modern terminology, if two ratios are "compounded" then they are multiplied together (translator's note). ↩