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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:

$$\frac{\operatorname{area}\boxdot{ABCD}}{\operatorname{area}\boxdot{CEFG}}=\frac{ab\sin(\gamma)}{a'b'\sin(\gamma')}=\frac{a}{a'}\cdot\frac{b}{b'}.$$

Table of Contents

Proofs: 1

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"

Adapted from CC BY-SA 3.0 Sources:

1. Prime.mover and others: "Pr∞fWiki", https://proofwiki.org/wiki/Main_Page, 2016

Footnotes