Page 391 - Euclid's Elements of Geometry

ST EW iþ.

ELEMENTS BOOK 10

δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ, ἐὰν ἄρα τῷ τετάρτῳ μέρει ously) laid down rational (straight-line) AC, and the
τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον square on the whole, AG, is greater than (the square on)
εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διελεῖ. τετμήσθω οὖν the attachment, DG, by the (square) on (some straight-
ἡ ΔΗ δίχα κατὰ τὸ Ε, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν line) commensurable (in length) with (AG) [Def. 10.13].
ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ Therefore, since the square on AG is greater than (the
ὑπὸ τῶν ΑΖ, ΖΗ. καὶ ἤχθωσαν διὰ τῶν Ε, Ζ, Η σημείων square on) GD by the (square) on (some straight-line)
τῇ ΑΓ παράλληλοι αἱ ΕΘ, ΖΙ, ΗΚ· σύμμετροι ἄρα εἰσὶν αἱ commensurable (in length) with (AG), thus if (an area)
ΑΖ, ΖΗ· σύμμετρον ἄρα καὶ τὸ ΑΙ τῷ ΖΚ. καὶ ἐπεὶ αἱ ΑΖ, equal to the fourth part of the square on DG is applied
ΖΗ σύμμετροί εἰσι μήκει, καὶ ἡ ΑΗ ἄρα ἑκατέρᾳ τῶν ΑΖ, to AG, falling short by a square figure, then it divides
ΖΗ σύμμετρός ἐστι μήκει. ῥητὴ δὲ ἡ ΑΗ καὶ ἀσύμμετρος (AG) into (parts which are) commensurable (in length)
τῇ ΑΓ μήκει· ὥστε καὶ αἱ ΑΖ, ΖΗ. ἑκάτερον ἄρα τῶν ΑΙ, [Prop. 10.17]. Therefore, let DG have been cut in half
ΖΚ μέσον ἐστίν. πάλιν, ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ at E. And let (an area) equal to the (square) on EG
μήκει, καὶ ἡ ΔΗ ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστι have been applied to AG, falling short by a square fig-
μήκει. ῥητὴ δὲ ἡ ΗΔ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· ῥητὴ ure. And let it be the (rectangle contained) by AF and
ἄρα καὶ ἑκατέρα τῶν ΔΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· FG. And let EH, FI, and GK have been drawn through
ἑκάτερον ἄρα τῶν ΔΘ, ΕΚ μέσον ἐστίν. καὶ ἐπεὶ αἱ ΑΗ, points E, F, and G (respectively), parallel to AC. Thus,
ΗΔ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα ἐστὶ AF and FG are commensurable (in length). AI (is) thus
μήκει ἡ ΑΗ τῇ ΗΔ. ἀλλ᾿ ἡ μὲν ΑΗ τῇ ΑΖ σύμμετρός ἐστι also commensurable with FK [Props. 6.1, 10.11]. And
μήκει ἡ δὲ ΔΗ τῇ ΕΗ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΕΗ since AF and FG are commensurable in length, AG is
μήκει. ὡς δὲ ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ thus also commensurable in length with each of AF and
ΕΚ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΙ τῷ ΕΚ. FG [Prop. 10.15]. And AG (is) rational, and incommen-
Συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ surable in length with AC. Hence, AF and FG (are)
δὲ ΖΚ ἴσον ἀφῇρήσθω τὸ ΝΞ περὶ τὴν αὐτὴν γωνίαν ὂν τῷ also (rational, and incommensurable in length with AC)
ΛΜ· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ. ἔστω [Prop. 10.13]. Thus, AI and FK are each medial (ar-
αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ eas) [Prop. 10.21]. Again, since DE is commensurable
οὖν τὸ ὑπὸ τῶν ΑΖ, ΖΗ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΗ, ἔστιν ἄρα in length with EG, DG is also commensurable in length
ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ. ἀλλ᾿ ὡς with each of DE and EG [Prop. 10.15]. And GD (is)
μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΕΚ· ὡς rational, and incommensurable in length with AC. Thus,
δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΖΚ· καὶ DE and EG (are) each also rational, and incommensu-
ὡς ἄρα τὸ ΑΙ πρὸς τὸ ΕΚ, οὕτως τὸ ΕΚ πρὸς τὸ ΖΚ· τῶν rable in length with AC [Prop. 10.13]. DH and EK are
ἄρα ΑΙ, ΖΚ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν thus each medial (areas) [Prop. 10.21]. And since AG
ΛΜ, ΝΞ τετραγώνων μέσον ἀνάλογον τὸ ΜΝ· καί ἐστιν and GD are commensurable in square only, AG is thus
ἴσον τὸ μὲν ΑΙ τῷ ΛΜ, τὸ δὲ ΖΚ τῷ ΝΞ· καὶ τὸ ΕΚ ἄρα incommensurable in length with GD. But, AG is com-
ἴσον ἐστὶ τῷ ΜΝ. ἀλλὰ τὸ μὲν ΜΝ ἴσον ἐστὶ τῷ ΛΞ, τὸ mensurable in length with AF, and DG with EG. Thus,
δὲ ΕΚ ἴσον [ἐστὶ] τῷ ΔΘ· καὶ ὅλον ἄρα τὸ ΔΚ ἴσον ἐστὶ AF is incommensurable in length with EG [Prop. 10.13].
τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἔστι δὲ καὶ τὸ ΑΚ ἴσον τοῖς And as AF (is) to EG, so AI is to EK [Prop. 6.1]. Thus,
ΛΜ, ΝΞ· λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΣΤ, τουτέστι τῷ AI is incommensurable with EK [Prop. 10.11].
ἀπὸ τῆς ΛΝ τετραγώνῳ· ἡ ΛΝ ἄρα δύναται τὸ ΑΒ χωρίον. Therefore, let the square LM, equal to AI, have
λέγω, ὅτι ἡ ΛΝ μέσης ἀποτομή ἐστι δευτέρα. been constructed. And let NO, equal to FK, which is
᾿Επεὶ γὰρ μέσα ἐδείχθη τὰ ΑΙ, ΖΚ καί ἐστιν ἴσα τοῖς ἀπὸ about the same angle as LM, have been subtracted (from
τῶν ΛΟ, ΟΝ, μέσον ἄρα καὶ ἑκάτερον τῶν ἀπὸ τῶν ΛΟ, LM). Thus, LM and NO are about the same diagonal
ΟΝ· μέση ἄρα ἑκατέρα τῶν ΛΟ, ΟΝ. καὶ ἐπεὶ σύμμετρόν [Prop. 6.26]. Let PR be their (common) diagonal, and
ἐστι τὸ ΑΙ τῷ ΖΚ, σύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τῷ let the (rest of the) figure have been drawn. Therefore,
ἀπὸ τῆς ΟΝ. πάλιν, ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΙ τῷ ΕΚ, since the (rectangle contained) by AF and FG is equal
ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΛΜ τῷ ΜΝ, τουτέστι τὸ ἀπὸ to the (square) on EG, thus as AF is to EG, so EG (is)
τῆς ΛΟ τῷ ὑπὸ τῶν ΛΟ, ΟΝ· ὥστε καὶ ἡ ΛΟ ἀσύμμετρός to FG [Prop. 6.17]. But, as AF (is) to EG, so AI is to
ἐστι μήκει τῇ ΟΝ· αἱ ΛΟ, ΟΝ ἄρα μέσαι εἰσὶ δυνάμει μόνον EK [Prop. 6.1]. And as EG (is) to FG, so EK is to FK
σύμμετροι. λέγω δή, ὅτι καὶ μέσον περιέχουσιν. [Prop. 6.1]. And thus as AI (is) to EK, so EK (is) to
᾿Επεὶ γὰρ μέσον ἐδείχθη τὸ ΕΚ καί ἐστιν ἴσον τῷ ὑπὸ FK [Prop. 5.11]. Thus, EK is the mean proportional to
τῶν ΛΟ, ΟΝ, μέσον ἄρα ἐστὶ καὶ τὸ ὑπὸ τῶν ΛΟ, ΟΝ· AI and FK. And MN is also the mean proportional to
ὥστε αἱ ΛΟ, ΟΝ μέσαι εἰσὶ δυνάμει μόνον σύμμετροι μέσον the squares LM and NO [Prop. 10.53 lem.]. And AI is

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