Euclid · Elements §12.prop.2#2

Proof of the Ratio of Circles and Related Lemma

Passage 270 of 316 · Greek

Summary

Concludes the proof by contradiction that the area of circles is proportional to the squares on their diameters. It also presents a lemma regarding the ratio between a larger area and a circle, which is used in the proof.

§12.prop.2#2ἀλλὰ καὶ ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸ Σ χωρίον· καὶ ὡς ἄρα ὁ ΑΒΓΔ κύκλος πρὸς τὸ Σ χωρίον, οὕτως τὸ ΑΞΒΟΓΠΔΡ πολύγωνον πρὸς τὸ ΕΚΖΛΗΜΘΝ πολύγωνον· ἐναλλὰξ ἄρα ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸ ἐν αὐτῷ πολύγωνον, οὕτως τὸ Σ χωρίον πρὸς τὸ ΕΚΖΛΗΜΘΝ πολύγωνον.

But also as the square on BD is to the square on ZQ, so is the circle ABGD to the area S; therefore also, as the circle ABGD is to the area S, so is the polygon AXBOGPDR to the polygon EKZLHMQN; therefore, alternately, as the circle ABGD is to the polygon in it, so is the area S to the polygon EKZLHMQN.

μείζων δὲ ὁ ΑΒΓΔ κύκλος τοῦ ἐν αὐτῷ πολυγώνου· μεῖζον ἄρα καὶ τὸ Σ χωρίον τοῦ ΕΚΖΛΗΜΘΝ πολυγώνου.

But the circle ABGD is greater than the polygon in it; therefore the area S is also greater than the polygon EKZLHMQN.

ἀλλὰ καὶ ἔλαττον· ὅπερ ἐστὶν ἀδύνατον.

But [the area S] is also less [than the polygon]; which is impossible.

οὐκ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς ἔλασσόν τι τοῦ ΕΖΗΘ κύκλου χωρίον.

Therefore, as the square on BD is to the square on ZQ, so is not the circle ABGD to some area less than the circle EZHQ.

ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ὡς τὸ ἀπὸ ΖΘ πρὸς τὸ ἀπὸ ΒΔ, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλασσόν τι τοῦ ΑΒΓΔ κύκλου χωρίον.

In the same way, indeed, we shall show that neither as the [square] on ZQ is to the [square] on BD, so is the circle EZHQ to some area less than the circle ABGD.

λέγω δή, ὅτι οὐδὲ ὡς τὸ ἀπὸ τῆς ΒΔ πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς μεῖζόν τι τοῦ ΕΖΗΘ κύκλου χωρίον.

I say indeed that neither, as the [square] on BD is to the [square] on ZQ, so is the circle ABGD to some area greater than the circle EZHQ.

εἰ γὰρ δυνατόν, ἔστω πρὸς μεῖζον τὸ Σ. ἀνάπαλιν ἄρα ὡς τὸ ἀπὸ τῆς ΖΘ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΔΒ, οὕτως τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον.

For, if possible, let it be to a greater [area], S. Therefore, conversely, as the square on ZQ is to the square on DB, so is the area S to the circle ABGD.

ἀλλʼ ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον· καὶ ὡς ἄρα τὸ ἀπὸ τῆς ΖΘ πρὸς τὸ ἀπὸ τῆς ΒΔ, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλασσόν τι τοῦ ΑΒΓΔ κύκλου χωρίον· ὅπερ ἀδύνατον ἐδείχθη.

But as the area S is to the circle ABGD, so is the circle EZHQ to some area less than the circle ABGD; therefore also, as the [square] on ZQ is to the [square] on BD, so is the circle EZHQ to some area less than the circle ABGD; which was proved impossible.

οὐκ ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς μεῖζόν τι τοῦ ΕΖΗΘ κύκλου χωρίον.

Therefore, as the square on BD is to the square on ZQ, so is not the circle ABGD to some area greater than the circle EZHQ.

ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἔλασσον· ἔστιν ἄρα ὡς τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ κύκλον.

And it was proved that neither is it to a less; therefore, as the square on BD is to the square on ZQ, so is the circle ABGD to the circle EZHQ.

οἱ ἄρα κύκλοι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ἀπὸ τῶν διαμέτρων τετράγωνα· ὅπερ ἔδει δεῖξαι.

Therefore, circles are to one another as the squares on their diameters; which it was required to prove.

λῆμμα λέγω δή, ὅτι τοῦ Σ χωρίου μείζονος ὄντος τοῦ ΕΖΗΘ κύκλου ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον.

Lemma I say indeed that, the area S being greater than the circle EZHQ, as the area S is to the circle ABGD, so is the circle EZHQ to some area less than the circle ABGD.

γεγονέτω γὰρ ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον.

For let it be that as the area S is to the circle ABGD, so is the circle EZHQ to the area T.

λέγω, ὅτι ἔλαττόν ἐστι τὸ Τ χωρίον τοῦ ΑΒΓΔ κύκλου.

I say that the area T is less than the circle ABGD.

ἐπεὶ γάρ ἐστιν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον, ἐναλλάξ ἐστιν ὡς τὸ Σ χωρίον πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως ὁ ΑΒΓΔ κύκλος πρὸς τὸ Τ χωρίον.

For since, as the area S is to the circle ABGD, so is the circle EZHQ to the area T, alternately, as the area S is to the circle EZHQ, so is the circle ABGD to the area T.

μεῖζον δὲ τὸ Σ χωρίον τοῦ ΕΖΗΘ κύκλου· μείζων ἄρα καὶ ὁ ΑΒΓΔ κύκλος τοῦ Τ χωρίου.

But the area S is greater than the circle EZHQ; therefore the circle ABGD is also greater than the area T.

ὥστε ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, οὕτως ὁ ΕΖΗΘ κύκλος πρὸς ἔλαττόν τι τοῦ ΑΒΓΔ κύκλου χωρίον· ὅπερ ἔδει δεῖξαι.

So that, as the area S is to the circle ABGD, so is the circle EZHQ to some area less than the circle ABGD; which it was required to prove.

Notes

¦60¦ἀλλὰ καὶ ἔλαττον — The subject 'the area S' (τὸ Σ χωρίον) and the object of comparison 'than the polygon EKZLHMQN' (τοῦ ΕΚΖΛΗΜΘΝ πολυγώνου) are omitted. This marks the contradiction, as it was established in the previous chunk that the polygon is greater than the area S (meaning the area S is less than the polygon).
¦85¦τοῦ Σ χωρίου μείζονος ὄντος τοῦ ΕΖΗΘ κύκλου — This is a genitive absolute construction (noun τοῦ Σ χωρίου + participle ὄντος + complement μείζονος), in which τοῦ ΕΖΗΘ κύκλου is a genitive of comparison ('than the circle EZHQ'). It expresses the assumption: 'the area S being greater than the circle EZHQ.'

Cite this passage

Euclid, Elements §12.prop.2#2. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:12.prop.2%232

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