# Proposition: 6.23: Ratio of Areas of Equiangular Parallelograms

### (Proposition 23 from Book 6 of Euclid's “Elements”)

Equiangular parallelograms have to one another the ratio compounded1 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.

### Modern Formulation

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

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