Page 336 - Euclid's Elements of Geometry

ST EW iþ.

√ nþ √ ELEMENTS BOOK 10
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whose length is k/ 1 − k ′ 2 − k [Prop. 10.86], are the roots of x − (2 k/ 1 − k ) x + k [k /(1 − k )] = 0.
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Proposition 50
Εὑρεῖν τὴν ἐκ δύο ὀνομάτων τρίτην. To find a third binomial (straight-line).
Α Γ Β A C B

Ε Κ ∆ E K D
Ζ Η Θ F G H

᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν Let the two numbers AC and CB be laid down such
συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον that their sum AB has to BC the ratio which (some)
ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, square number (has) to (some) square number, and does
πρὸς δὲ τὸν ΑΓ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς not have to AC the ratio which (some) square number
πρὸς τετράγωνον ἀριθμόν. ἐκκείσθω δέ τις καὶ ἄλλος μὴ (has) to (some) square number. And let some other non-
τετράγωνος ἀριθμὸς ὁ Δ, καὶ πρὸς ἑκάτερον τῶν ΒΑ, ΑΓ square number D also be laid down, and let it not have
λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον to each of BA and AC the ratio which (some) square
ἀριθμόν· καὶ ἐκκείσθω τις ῥητὴ εὐθεῖα ἡ Ε, καὶ γεγονέτω number (has) to (some) square number. And let some ra-
ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ tional straight-line E be laid down, and let it have been
τῆς ΖΗ· σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς Ε τῷ ἀπὸ τῆς ΖΗ. contrived that as D (is) to AB, so the (square) on E
καί ἐστι ῥητὴ ἡ Ε· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ Δ (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the
πρὸς τὸν ΑΒ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς (square) on E is commensurable with the (square) on
τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς FG [Prop. 10.6]. And E is a rational (straight-line).
ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον Thus, FG is also a rational (straight-line). And since
ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ ΖΗ μήκει. γεγονέτω D does not have to AB the ratio which (some) square
δὴ πάλιν ὡς ἡ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς number has to (some) square number, the (square) on
ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ· σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς E does not have to the (square) on FG the ratio which
ΖΗ τῷ ἀπὸ τῆς ΗΘ. ῥητὴ δὲ ἡ ΖΗ· ῥητὴ ἄρα καὶ ἡ ΗΘ. καὶ (some) square number (has) to (some) square number
ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος either. E is thus incommensurable in length with FG
ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΖΗ [Prop. 10.9]. So, again, let it have been contrived that
πρὸς τὸ ἀπὸ τῆς ΘΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς as the number BA (is) to AC, so the (square) on FG
πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ (is) to the (square) on GH [Prop. 10.6 corr.]. Thus, the
τῇ ΗΘ μήκει. αἱ ΖΗ, ΗΘ ἄρα ῥηταί εἰσι δυνάμει μόνον (square) on FG is commensurable with the (square) on
σύμμετροι· ἡ ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστίν. λέγω δή, ὅτι GH [Prop. 10.6]. And FG (is) a rational (straight-line).
καὶ τρίτη. Thus, GH (is) also a rational (straight-line). And since
᾿Επεὶ γάρ ἐστιν ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς BA does not have to AC the ratio which (some) square
Ε πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως number (has) to (some) square number, the (square) on
τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, δι᾿ ἴσου ἄρα ἐστὶν ὡς FG does not have to the (square) on HG the ratio which
ὁ Δ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς (some) square number (has) to (some) square number
ΗΘ. ὁ δὲ Δ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος either. Thus, FG is incommensurable in length with GH
ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα [Prop. 10.9]. FG and GH are thus rational (straight-
πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς lines which are) commensurable in square only. Thus,
πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ FH is a binomial (straight-line) [Prop. 10.36]. So, I say
ΗΘ μήκει. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως that (it is) also a third (binomial straight-line).
τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, μεῖζον ἄρα τὸ ἀπὸ For since as D is to AB, so the (square) on E (is)
τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ. ἔστω οὖν τῷ ἀπὸ τῆς ΖΗ ἴσα τὰ to the (square) on FG, and as BA (is) to AC, so the
ἀπὸ τῶν ΗΘ, Κ· ἀναστρέψαντι ἄρα [ἐστὶν] ὡς ὁ ΑΒ πρὸς (square) on FG (is) to the (square) on GH, thus, via
τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ equality, as D (is) to AC, so the (square) on E (is)
ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς to the (square) on GH [Prop. 5.22]. And D does not

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