§6.33λγ΄.
33.
Κύκλος περὶ διάμετρον τὴν ΑΒ, καὶ ἐκβεβλήσθω ἡ ΑΒ καὶ ἔστω ἐπὶ τυχοῦσαν τὴν ∠Ε κάθετος, καὶ τῷ ὑπὸ ΑΖΒ ἴσον κείσθω τὸ ἀπὸ ΖΗ τετράγωνον· ὅτι, οἷον ἐὰν ληφθῇ σημεῖον ὡς τὸ Ε, καὶ ἀπ᾿ αὐτοῦ ἐπὶ τὸ Η ἐπιζευχθεῖσα ἐκβληθῇ ἐπὶ τὸ Θ, γίνεται καὶ τὸ ὑπὸ ΘΕΚ ἴσον τῷ ἀπὸ ΚΗ τετραγώνῳ.
Let there be a circle about the diameter ΑΒ, and let ΑΒ be produced, and let a perpendicular be drawn to any (straight line) ∠Ε, and let the square on ΖΗ be made equal to the rectangle contained by ΑΖ, ΖΒ; (to prove) that, if any point such as Ε is taken on ΕΔ, and the line joined from it to Η is produced to Θ, the rectangle contained by ΘΕ, ΕΚ also becomes equal to the square on ΚΗ.
ἐπεζεύχθωσαν αἱ ΑΕ, ΒΛ ὀρθὴ ἄρα ἐστὶν ἡ Θ Λ γωνία.
Let ΑΕ, ΒΛ be joined; therefore angle ΘΛ[Β] is a right angle.
ἔστιν δὲ καὶ ἡ Ζ ὀρθή· τὸ ἄρα ὑπὸ ΑΕΛ ἴσον ἐστὶν τῷ τε ὑπὸ ΑΖΒ καὶ τῷ ἀπὸ ΖΕ τετραγώνῳ.
And Ζ is also a right angle; therefore the rectangle contained by ΑΕ, ΕΛ is equal to both the rectangle contained by ΑΖ, ΖΒ and the square on ΖΕ.
ἀλλὰ τὸ μὲν ὑπὸ ΑΕΛ ἴσον ἐστὶν τῷ ὑπὸ ΘΕΚ, τὸ δὲ ὑπὸ ΑΖΒ ἴσον ἐστὶν τῷ ἀπὸ τετραγώνῳ· τὸ ἄρα ὑπὸ ΘΕΚ ἴσον ἐστὶν τοῖς ἀπὸ τῶν ΕΖ, ΖΗ τετραγώνοις, τουτέστιν τῷ ἀπὸ ΕΗ τετραγώνῳ.
But the rectangle contained by ΑΕ, ΕΛ is equal to the rectangle contained by ΘΕ, ΕΚ, and the rectangle contained by ΑΖ, ΖΒ is equal to the square on [ΖΗ]; therefore the rectangle contained by ΘΕ, ΕΚ is equal to the squares on ΕΖ, ΖΗ, that is, to the square on ΕΗ.
§6.34λδ΄.
34.
Ἔστω, ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΑΔ πρὸς τὴν ∠Γ. καὶ τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε σημεῖον· ὅτι γίνεται τρία, τὸ μὲν ὑπὸ ΒΕ∠ ἴσον τῷ ἀπὸ ΕΓ τετραγώνῳ, τὸ δὲ ὑπὸ Β∠Ε τῷ ὑπὸ Α∠Γ, τὸ δὲ ὑπὸ ΑΒΓ τῷ ὑπὸ ΕΒ∠.
Let it be that, as ΑΒ is to ΒΓ, so is ΑΔ to ∠Γ. And let ΑΓ be bisected at the point Ε; (to prove) that three things result: first, the rectangle contained by ΒΕ, ΕΔ is equal to the square on ΕΓ; second, the rectangle contained by ΒΔ, ΔΕ is equal to the rectangle contained by ΑΔ, ΔΓ; third, the rectangle contained by ΑΒ, ΒΓ is equal to the rectangle contained by ΕΒ, ΒΔ.
ἐπεὶ γάρ, ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ Α∠ πρὸς τὴν ∠Γ, συνθέντι καὶ τὰ ἡμίση τῶν ἡγουμένων καὶ ἀναστρέψαντι ἄρα ἐστίν, ὡς ἡ ΒΕ πρὸς τὴν ΕΓ, ἡ ΕΓ πρὸς τὴν Ε∠· τὸ ἄρα ὑπὸ ΒΕ∠ ἴσον ἐστὶν τῷ ἀπὸ ΕΓ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ ΔΕ τετράγωνον· λοιπὸν ἄρα τὸ ὑπὸ Β∠Ε ἴσον ἐστὶν τῷ ὑπὸ Α∠Γ. πάλιν τὸ ὑπὸ ΒΕ∠ ἴσον ἐστὶν τῷ ἀπὸ ΕΓ τετραγώνῳ.
For since, as ΑΒ is to ΒΓ, so is ΑΔ to ΔΓ, by compounding, taking the halves of the antecedents, and converting, it is therefore as ΒΕ is to ΕΓ, so is ΕΓ to ΕΔ; therefore the rectangle contained by ΒΕ, ΕΔ is equal to the square on ΕΓ. Let the common square on ΔΕ be subtracted; therefore the remaining rectangle contained by ΒΔ, ΔΕ is equal to the rectangle contained by ΑΔ, ΔΓ. Again, the rectangle contained by ΒΕ, ΕΔ is equal to the square on ΕΓ.
ἀμφότερα ἀφῃρήσθω ἀπὸ τοῦ ἀπὸ τῆς ΒΕ τετραγώνου· λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶν τῷ ὑπὸ τῶν ΕΒ∠.
Let both be subtracted from the square on ΒΕ; therefore the remaining rectangle contained by ΑΒ, ΒΓ is equal to the rectangle contained by ΕΒ, ΒΔ.
Ἀλλὰ ἔστω νῦν τὸ ὑπὸ τῶν Β∠Ε ἴσον τῷ ὑπὸ τῶν Α∠Γ, καὶ τετμήσθω δίχα ἡ ΓΑ κατὰ τὸ Ε· ὅτι ἐστίν, ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ Α∠ πρὸς τὴν ΔΓ.
ἐπεὶ γὰρ τὸ ὑπὸ τῶν Β∠Ε ἴσον ἐστὶν τῷ ὑπὸ τῶν Α∠Γ, κοινὸν προσκείσθω τὸ ἀπὸ ∠Ε τετράγωνον· ὅλον ἄρα τὸ ὑπὸ ΒΕ∠ ἴσον τῷ ἀπὸ ΓΕ τετραγώνῳ.
But let it now be that the rectangle contained by ΒΔ, ΔΕ is equal to the rectangle contained by ΑΔ, ΔΓ, and let ΓΑ be bisected at Ε; (to prove) that, as ΑΒ is to ΒΓ, so is ΑΔ to ΔΓ. For since the rectangle contained by ΒΔ, ΔΕ is equal to the rectangle contained by ΑΔ, ΔΓ, let the common square on ΔΕ be added; therefore the whole rectangle contained by ΒΕ, ΕΔ is equal to the square on ΓΕ.
ἀνάλογον καὶ ἀναστρέφοντι καὶ δὶς τὰ ἡγούμενα καὶ διελόντι ἄρα ἐστίν, ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ Α∠ πρὸς τὴν ∠Γ.
Standardly, and by converting, doubling the antecedents, and separating, it is therefore as ΑΒ is to ΒΓ, so is ΑΔ to ΔΓ.
§6.35λε΄.
35.
Τούτων ὄντων ἔστω κύκλος ὁ περὶ διάμετρον τὴν ΑΒ, καὶ ἐκβεβλήσθω ἡ ΑΒ, ἔστω δὲ ἐπὶ τυχοῦσαν τὴν ∠Ε κάθετος, καὶ πεποιήσθω, ὡς ἡ ΑΝ πρὸς τὴν ΖΒ, οὕτως ἡ ΑΗ πρὸς τὴν ΗΒ ὅτι πάλιν, οἷον ἐὰν ἐπὶ τῆς ΕΔ σημεῖον ληφθῇ ὡς τὸ Ε, καὶ ἐπιζευχθεῖσα ἡ ΕΗ ἐκβληθῇ ἐπὶ τὸ Θ, γίνεται, ὡς ἡ ΘΕ πρὸς τὴν ΕΚ, οὕτως ἡ ΘΗ πρὸς τὴν ΗΚ.
εἰλήφθω τὸ κέντρον τοῦ κύκλου τὸ Λ, καὶ ἀπὸ τοῦ Λ ἐπὶ τὴν ΕΘ κάθετος ἤχθω ἡ ΛΜ· ἴση ἄρα ἐστὶν ἡ ΚΜ τῇ ΜΘ. ἐπεὶ δὲ ὀρθή ἔστιν ἑκατέρα τῶν Μ, Ζ γωνιῶν, ἐν κύκλῳ ἐστὶν τὰ Ε, Ζ, Λ, Μ σημεῖα· τὸ ἄρα ὑπὸ ΖΗΛ ἴσον ἐστὶν τῷ ὑπὸ τῶν ΕΗΜ. ἀλλὰ τὸ ὑπὸ τῶν ΖΗΛ ἴσον ἐστὶν τῷ ὑπὸ τῶν ΑΗΒ διὰ τὸ εἶναι, ὡς τὴν ΑΖ πρὸς τὴν ΖΒ, οὕτως τὴν ΑΗ πρὸς τὴν ΗΒ, καὶ τετμῆσθαι τὴν ΑΒ δίχα κατὰ τὸ Λ· καὶ τὸ ὑπὸ τῶν ΕΗΜ ἄρα ἴσον ἐστὶν τῷ ὑπὸ τῶν ΑΗΒ, τουτέστιν· ἐν κύκλῳ γάρ· τῷ ὑπὸ τῶν ΘΗΚ. καὶ τέτμηται δίχα ἡ ΘΚ κατὰ τὸ Μ· διὰ δὴ τὸ προγεγραμμένον γίνεται, ὡς ἡ ΘΕ πρὸς τὴν ΕΚ, οὕτως ἡ ΘΗ πρὸς τὴν ΗΚ.
These things being so, let there be a circle about the diameter ΑΒ, and let ΑΒ be produced, and let a perpendicular be drawn to any (straight line) ∠Ε, and let it be made that, as Α[Ζ] is to ΖΒ, so is ΑΗ to ΗΒ; (to prove) that again, if any point such as Ε is taken on ΕΔ, and ΕΗ being joined is produced to Θ, it results that, as ΘΕ is to ΕΚ, so is ΘΗ to ΗΚ. Let the center of the circle be taken as Λ, and from Λ let the perpendicular ΛΜ be drawn to ΕΘ; therefore ΚΜ is equal to ΜΘ. Since each of the angles Μ, Ζ is a right angle, the points Ε, Ζ, Λ, Μ are on a circle; therefore the rectangle contained by ΖΗ, ΗΛ is equal to the rectangle contained by ΕΗ, ΗΜ. But the rectangle contained by ΖΗ, ΗΛ is equal to the rectangle contained by ΑΗ, ΗΒ, because it is, as ΑΖ is to ΖΒ, so is ΑΗ to ΗΒ, and ΑΒ is bisected at Λ; therefore the rectangle contained by ΕΗ, ΗΜ is also equal to the rectangle contained by ΑΗ, ΗΒ, that is—for they are on a circle—to the rectangle contained by ΘΗ, ΗΚ. And ΘΚ is bisected at Μ; therefore by what was written before, as ΘΕ is to ΕΚ, so is ΘΗ to ΗΚ.