Areas of Circles are as Squares on Diameters/Lemma
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Theorem
In the words of Euclid:
- I say that, the area $S$ being greater than the circle $EFGH$, as the area $S$ is to the circle $ABCD$, so is the circle $EFGH$ to some area less than the circle $ABCD$.
(The Elements: Book $\text{XII}$: Proposition $2$ : Lemma)
Proof
Let it be contrived that:
- $S : ABCD = EFGH : T$
for some area $T$.
It is to be demonstrated that $T$ is less than the circle $ABCD$.
From the relationship:
- $S : ABCD = EFGH : T$
it follows from Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately that:
- $S : EFGH = ABCD : T$
But:
- $S > EFGH$
Therefore:
- $ABCD > T$
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $2$ of Book $\text{XII}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XII}$. Propositions