§6.prop.29παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ὑπερβάλλον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ δοθέντι.
To apply to a given straight line a parallelogram equal to a given rectilineal figure, exceeding by a parallelogrammic figure similar to a given one.
ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν εὐθύγραμμον, ᾧ δεῖ ἴσον παρὰ τὴν ΑΒ παραβαλεῖν, τὸ Γ, ᾧ δὲ δεῖ ὅμοιον ὑπερβάλλειν, τὸ Δ· δεῖ δὴ παρὰ τὴν ΑΒ εὐθεῖαν τῷ Γ εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ὑπερβάλλον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ Δ.
τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε, καὶ ἀναγεγράφθω ἀπὸ τῆς ΕΒ τῷ Δ ὅμοιον καὶ ὁμοίως κείμενον παραλληλόγραμμον τὸ ΒΖ, καὶ συναμφοτέροις μὲν τοῖς ΒΖ, Γ ἴσον, τῷ δὲ Δ ὅμοιον καὶ ὁμοίως κείμενον τὸ αὐτὸ συνεστάτω τὸ ΗΘ. ὁμόλογος δὲ ἔστω ἡ μὲν ΚΘ τῇ ΖΛ, ἡ δὲ ΚΗ τῇ ΖΕ. καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΗΘ τοῦ ΖΒ, μείζων ἄρα ἐστὶ καὶ ἡ μὲν ΚΘ τῆς ΖΛ, ἡ δὲ ΚΗ τῆς ΖΕ. ἐκβεβλήσθωσαν αἱ ΖΛ, ΖΕ, καὶ τῇ μὲν ΚΘ ἴση ἔστω ἡ ΖΛΜ, τῇ δὲ ΚΗ ἴση ἡ ΖΕΝ, καὶ συμπεπληρώσθω τὸ ΜΝ·
Let the given straight line be ΑΒ, and the given rectilineal figure, to which the parallelogram to be applied to ΑΒ must be equal, Γ, and that to which the excess must be similar, Δ; it is required then to apply to the straight line ΑΒ a parallelogram equal to the rectilineal figure Γ, exceeding by a parallelogrammic figure similar to Δ.
τὸ ΜΝ ἄρα τῷ ΗΘ ἴσον τέ ἐστι καὶ ὅμοιον.
Let ΑΒ be bisected at Ε, and let there be described on ΕΒ a parallelogram ΒΖ similar and similarly situated to Δ, and let ΗΘ be constructed equal to both ΒΖ and Γ together, and similar and similarly situated to Δ. And let ΚΘ correspond to ΖΛ, and ΚΗ to ΖΕ. And, since ΗΘ is greater than ΖΒ, therefore ΚΘ is also greater than ΖΛ, and ΚΗ than ΖΕ. Let ΖΛ and ΖΕ be produced, and let ΖΛΜ be equal to ΚΘ, and ΖΕΝ equal to ΚΗ, and let ΜΝ be completed; therefore ΜΝ is equal and similar to ΗΘ.
ἀλλὰ τὸ ΗΘ τῷ ΕΛ ἐστιν ὅμοιον· καὶ τὸ ΜΝ ἄρα τῷ ΕΛ ὅμοιόν ἐστιν· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὸ ΕΛ τῷ ΜΝ. ἤχθω αὐτῶν διάμετρος ἡ ΖΞ, καὶ καταγεγράφθω τὸ σχῆμα.
But ΗΘ is similar to ΕΛ; therefore ΜΝ is also similar to ΕΛ; therefore ΕΛ is about the same diameter with ΜΝ. Let ΖΞ be their diameter, and let the figure be described.
ἐπεὶ ἴσον ἐστὶ τὸ ΗΘ τοῖς ΕΛ, Γ, ἀλλὰ τὸ ΗΘ τῷ ΜΝ ἴσον ἐστίν, καὶ τὸ ΜΝ ἄρα τοῖς ΕΛ, Γ ἴσον ἐστίν.
Since ΗΘ is equal to ΕΛ and Γ together, but ΗΘ is equal to ΜΝ, therefore ΜΝ is also equal to ΕΛ and Γ together.
κοινὸν ἀφῃρήσθω τὸ ΕΛ· λοιπὸς ἄρα ὁ ΨΧΦ γνώμων τῷ Γ ἐστιν ἴσος.
Let ΕΛ be subtracted as common; therefore the remainder, the gnomon ΨΧΦ, is equal to Γ.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, ἴσον ἐστὶ καὶ τὸ ΑΝ τῷ ΝΒ, τουτέστι τῷ ΛΟ. κοινὸν προσκείσθω τὸ ΕΞ· ὅλον ἄρα τὸ ΑΞ ἴσον ἐστὶ τῷ ΦΧΨ γνώμονι.
And since ΑΕ is equal to ΕΒ, ΑΝ is also equal to ΝΒ, that is, to ΛΟ. Let ΕΞ be added as common; therefore the whole ΑΞ is equal to the gnomon ΦΧΨ.
ἀλλὰ ὁ ΦΧΨ γνώμων τῷ Γ ἴσος ἐστίν· καὶ τὸ ΑΞ ἄρα τῷ Γ ἴσον ἐστίν.
But the gnomon ΦΧΨ is equal to Γ; therefore ΑΞ is also equal to Γ.
παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΑΞ ὑπερβάλλον εἴδει παραλληλογράμμῳ τῷ ΠΟ ὁμοίῳ ὄντι τῷ Δ, ἐπεὶ καὶ τῷ ΕΛ ἐστιν ὅμοιον τὸ ΟΠ· ὅπερ ἔδει ποιῆσαι.
Therefore, to the given straight line ΑΒ there has been applied a parallelogram ΑΞ equal to the given rectilineal figure Γ, exceeding by a parallelogrammic figure ΠΟ similar to Δ, since ΟΠ is also similar to ΕΛ; which was to be done.
§6.prop.30τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην ἄκρον καὶ μέσον λόγον τεμεῖν.
To cut a given finite straight line in extreme and mean ratio.
ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ· δεῖ δὴ τὴν ΑΒ εὐθεῖαν ἄκρον καὶ μέσον λόγον τεμεῖν.
Let the given finite straight line be ΑΒ; it is required then to cut the straight line ΑΒ in extreme and mean ratio.
Ἀναγεγράφθω ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΒΓ, καὶ παραβεβλήσθω παρὰ τὴν ΑΓ τῇ ΒΓ ἴσον παραλληλόγραμμον τὸ ΓΔ ὑπερβάλλον εἴδει τῷ ΑΔ ὁμοίῳ τῷ ΒΓ.
τετράγωνον δέ ἐστι τὸ ΒΓ· τετράγωνον ἄρα ἐστὶ καὶ τὸ ΑΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΒΓ τῷ ΓΔ, κοινὸν ἀφῃρήσθω τὸ ΓΕ· λοιπὸν ἄρα τὸ ΒΖ λοιπῷ τῷ ΑΔ ἐστιν ἴσον.
Let there be described on ΑΒ the square ΒΓ, and let there be applied to ΑΓ a parallelogram ΓΔ equal to ΒΓ, exceeding by a figure ΑΔ similar to ΒΓ. And ΒΓ is a square; therefore ΑΔ is also a square. And since ΒΓ is equal to ΓΔ, let ΓΕ be subtracted as common; therefore the remainder ΒΖ is equal to the remainder ΑΔ.
ἔστι δὲ αὐτῷ καὶ ἰσογώνιον· τῶν ΒΖ, ΑΔ ἄρα ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· ἔστιν ἄρα ὡς ἡ ΖΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΑΕ πρὸς τὴν ΕΒ. ἴση δὲ ἡ μὲν ΖΕ τῇ ΑΒ, ἡ δὲ ΕΔ τῇ ΑΕ. ἔστιν ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΕ, οὕτως ἡ ΑΕ πρὸς τὴν ΕΒ. μείζων δὲ ἡ ΑΒ τῆς ΑΕ· μείζων ἄρα καὶ ἡ ΑΕ τῆς ΕΒ.
ἡ ἄρα ΑΒ εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ε, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστι τὸ ΑΕ· ὅπερ ἔδει ποιῆσαι.
And it is also equiangular with it; therefore in ΒΖ and ΑΔ the sides about the equal angles are reciprocally proportional; therefore, as ΖΕ is to ΕΔ, so is ΑΕ to ΕΒ. But ΖΕ is equal to ΑΒ, and ΕΔ to ΑΕ. Therefore, as ΒΑ is to ΑΕ, so is ΑΕ to ΕΒ. And ΑΒ is greater than ΑΕ; therefore ΑΕ is also greater than ΕΒ. Therefore the straight line ΑΒ has been cut in extreme and mean ratio at Ε, and the greater segment of it is ΑΕ; which was to be done.