Humanitext Reader

Euclid · Fragments §6.26-6.28

Equivalence of Squares from Segment Ratios and Secants

Passage 17 of 29 · Greek

Summary

In Theorems 26 and 27, the equality between a rectangle and a square is proven from the ratio of line segments and their squares. In Theorem 28, a property of line ratios regarding tangents and secants of a circle is proven using the equivalence of rectangles and squares.

§6.26κϚ΄.
26.
Ἔστω, ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ Α∠ πρὸς τὸ ἀπὸ ∠ ὅτι τὸ ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶν τῷ ἀπὸ τῆς Β∠ τετραγώνῳ.
Let, as ΑΒ is to ΒΓ, so is the square on Α∠ to the square on ∠; (to prove) that the rectangle contained by ΑΒ, ΒΓ is equal to the square on Β∠.
κείσθω τῇ Γ∠ ἴση ἡ ∠Ε· τὸ ἄρα ὑπὸ ΕΑΓ μετὰ τοῦ ἀπὸ Γ∠, τουτέστιν τοῦ ὑπὸ Γ∠Ε, ἴσον τῷ ἀπὸ Α∠.
Let ∠Ε be placed equal to Γ∠; therefore, the rectangle contained by ΕΑ, ΑΓ together with the square on Γ∠, that is, the rectangle contained by Γ∠, ∠Ε, is equal to the square on Α∠.
ἐπεὶ οὗν ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ Α∠ πρὸς τὸ ἀπὸ ∠Γ, διελόντι ἐστίν, ὡς ἡ ΑΓ πρὸς τὴν ΓΒ, τουτέστιν ὡς τὸ ὑπὸ ΕΑΓ πρὸς τὸ ὑπὸ ΕΑ, ΒΓ οὕτως τὸ ὑπὸ ΕΑΓ πρὸς τὸ ὑπὸ Γ∠Ε ἴσον ἄρα ἐστὶν τὸ ὑπὸ ΑΕ, ΒΓ τῷ ὑπὸ Γ∠Ε. ἀνάλογον καὶ διελόντι ἐστίν, ὡς ἡ Α∠ πρὸς τὴν ∠Ε, τουτέστιν πρὸς τὴν ∠Γ, οὕτως ἡ ∠Β πρὸς τὴν ΒΓ καὶ λοιπὴ ἄρα ἡ ΑΒ πρὸς λοιπὴν τὴν Β∠ ἐστιν, ὡς ἡ Β∠ πρὸς τὴν ΒΓ. τὸ ἄρα ὑπὸ ΑΒΓ ἴσον ἐστὶν τῷ ἀπὸ τῆς Β∠ τετραγώνῳ.
Since, therefore, as ΑΒ is to ΒΓ, so is the square on Α∠ to the square on ∠Γ, by subtraction, as ΑΓ is to ΓΒ, that is, as the rectangle contained by ΕΑ, ΑΓ is to the rectangle contained by ΕΑ, ΒΓ, so is the rectangle contained by ΕΑ, ΑΓ to the rectangle contained by Γ∠, ∠Ε; therefore, the rectangle contained by ΑΕ, ΒΓ is equal to the rectangle contained by Γ∠, ∠Ε. Proportionally and by subtraction, as Α∠ is to ∠Ε, that is, to ∠Γ, so is ∠Β to ΒΓ; therefore, also the remaining ΑΒ is to the remaining Β∠, as Β∠ is to ΒΓ. Therefore, the rectangle contained by ΑΒ, ΒΓ is equal to the square on Β∠.
§6.27κζ΄.
27.
Ἔστω δὲ πάλιν, ὡς ἡ ΑΒ πρὸς τὴν ΒΓ. οὕτως τὸ ἀπὸ Α∠ τετράγωνον πρὸς τὸ ἀπὸ ∠Γ τετράγωνον· ὅτι τὸ ὑπὸ ΑΒΓ ἴσον ἐστὶν τῷ ἀπὸ τῆς Β∠ τετραγώνῳ.
And let again, as ΑΒ is to ΒΓ, so is the square on Α∠ to the square on ∠Γ; (to prove) that the rectangle contained by ΑΒ, ΒΓ is equal to the square on Β∠.
κείσθω γὰρ ὁμοίως τῇ Γ∠ ἴση ἡ ∠Ε· τὸ ἄρα ὑπὸ ΓΑΕ μετὰ τοῦ ἀπὸ Γ∠, τουτέστιν τοῦ ὑπὸ Ε∠Γ, ἴσον τῷ ἀπὸ Α∠.
For let ∠Ε be placed likewise equal to Γ∠; therefore, the rectangle contained by ΓΑ, ΑΕ together with the square on Γ∠, that is, the rectangle contained by Ε∠, ∠Γ, is equal to the square on Α∠.
καὶ γίνεται κατὰ διαίρεσιν, ὡς ἡ ΑΓ πρὸς τὴν ΓΒ, τουτέστιν ὡς τὸ ὑπὸ ΕΑΓ πρὸς τὸ ὑπὸ ΕΑ, ΓΒ, οὕτως τὸ ὑπὸ ΓΑΕ πρὸς τὸ ὑπὸ Ε∠ ἴσον ἄρα ἐστὶν τὸ ὑπὸ ΑΕ. ΓΒ τῷ ὑπὸ Ε∠Γ. ἀνάλογον καὶ συνθέντι ἐστίν, ὡς ἡ Α∠ πρὸς τὴν ∠Ε, τουτέστιν πρὸς τὴν ∠Γ. οὕτως ἡ ∠Β πρὸς τὴν ΒΓ καὶ ὅλη ἄρα ἡ ΑΒ πρὸς ὅλην τὴν Β∠ ἐστιν, ὡς ἡ Β∠ πρὸς τὴν ΒΓ. τὸ ἄρα ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶν τῷ ἀπὸ τῆς Β∠ τετραγώνῳ.
And it becomes by division, as ΑΓ is to ΓΒ, that is, as the rectangle contained by ΕΑ, ΑΓ is to the rectangle contained by ΕΑ, ΓΒ, so is the rectangle contained by ΓΑ, ΑΕ to the rectangle contained by Ε∠; therefore, the rectangle contained by ΑΕ, ΓΒ is equal to the rectangle contained by Ε∠, ∠Γ. Proportionally and by addition, as Α∠ is to ∠Ε, that is, to ∠Γ, so is ∠Β to ΒΓ; therefore, also the whole ΑΒ is to the whole Β∠, as Β∠ is to ΒΓ. Therefore, the rectangle contained by ΑΒ, ΒΓ is equal to the square on Β∠.
§6.28κη΄.
28.
Κύκλου τοῦ ΑΒΓ ἐφαπτέσθωσαν αἱ Α∠, ∠Γ, καὶ ἐπεζεύχθω ἡ ΑΓ. καὶ διήχθω τυχοῦσα ἡ ∠Β·
Let Α∠, ∠Γ be tangent to the circle ΑΒΓ, and let ΑΓ be joined.
ὅτι γίνεται, ὡς ἡ Β∠ πρὸς τὴν ∠Ε, οὕτως ἡ Β πρὸς τὴν ΖΕ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ Α∠ τῇ ∠Γ, τὸ ἄρα ὑπὸ ΑΖΓ μετὰ τοῦ ἀπὸ Ζ∠ ἴσον ἐστὶν τῷ ἀπὸ ∠Α. ἀλλὰ τὸ μὲν ὑπὸ ΑΖΓ ἴσον ἐστὶν τῷ ὑπὸ ΒΖΕ, τὸ δὲ ἀπὸ ∠Α. ἐστιν τὸ ὑπὸ Β∠Ε·
And let an arbitrary line ∠Β be drawn; (to prove) that, as Β∠ is to ∠Ε, so is Β [Ζ] to ΖΕ. For since Α∠ is equal to ∠Γ, therefore the rectangle contained by ΑΖ, ΖΓ together with the square on ∠Ζ is equal to the square on ∠Α. But the rectangle contained by ΑΖ, ΖΓ is equal to the rectangle contained by ΒΖ, ΖΕ, and the square on ∠Α is equal to the rectangle contained by Β∠, ∠Ε; therefore, the rectangle contained by ΒΖ, ΖΕ together with the square on ∠Ζ is equal to the rectangle contained by Β∠, ∠Ε.
τὸ ἄρα ὑπὸ ΒΖΕ μετὰ τοῦ ἀπὸ ∠Ζ ἴσον ἐστὶν τῷ ὑπὸ Β∠Ε. ἐὰν δὲ τοῦτο, γίνεται, ὡς ἡ Β∠ πρὸς τὴν ∠Ε, οὕτως ἡ ΒΖ πρὸς τὴν ΖΕ.
And if this is so, it becomes, as Β∠ is to ∠Ε, so is ΒΖ to ΖΕ.

Notes

  1. 6.26τὸ ἀπὸ ∠ — The expression "τὸ ἀπὸ ∠" in the text lacks "Γ" and actually refers to "the square on ∠Γ" (τὸ ἀπὸ ∠Γ), as confirmed by the context and the subsequent steps of the proof.
  2. 6.26τοῦ ὑπὸ Γ∠Ε — In "τοῦ ὑπὸ Γ∠Ε", this refers to the rectangle contained by the two segments Γ∠ and ∠Ε. In geometric texts, such a shorthand with three letters is often used to designate the rectangle contained by two adjacent segments.
  3. 6.27τὸ ὑπὸ Ε∠ — In "τὸ ὑπὸ Ε∠", considering the correspondence with the preceding "τοῦ ὑπὸ Ε∠Γ" (the rectangle contained by Ε∠, ∠Γ), the letter "Γ" has been omitted, and it is understood as "the rectangle contained by Ε∠ and ∠Γ".
  4. 6.28οὕτως ἡ Β πρὸς τὴν ΖΕ — In "ἡ Β πρὸς τὴν ΖΕ", comparing it with "οὕτως ἡ ΒΖ πρὸς τὴν ΖΕ" at the end of the proof, it is clear that the letter "Ζ" has been omitted after "Β", and it actually refers to "ΒΖ".
  5. 6.28ἐὰν δὲ τοῦτο — The elliptical expression "ἐὰν δὲ τοῦτο" (if this is so) indicates that if the preceding equality (the rectangle ΒΖ, ΖΕ plus the square on ∠Ζ equals the rectangle Β∠, ∠Ε) holds, then the subsequent proportion is established.

Cite this passage

Euclid, Fragments §6.26-6.28. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg016.humanitext-grc1:6.26-6.28

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