Page 396 - Euclid's Elements of Geometry

ST EW iþ.

ELEMENTS BOOK 10

αὐτῶν σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ μήκει, ἡ GD are rational (straight-lines which are) commensu-
δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται rable in square only [Prop. 10.73], and neither of them is
τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ commensurable in length with the (previously) laid down
μεῖζον δύναται τῷ ἀπὸ ἁσυμμέτρου ἐαυτῇ μήκει, ἑὰν ἄρα rational (straight-line) AC, and the square on the whole,
τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ πα- AG, is greater than (the square on) the attachment, DG,
ραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν by the (square) on (some straight-line) incommensurable
διελεῖ. τετμήσθω οὖν ἡ ΔΗ δίχα κατὰ τὸ Ε [σημεῖον], καὶ in length with (AG) [Def. 10.16]. Therefore, since the
τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον square on AG is greater than (the square on) GD by
εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ· ἀσύμμετρος the (square) on (some straight-line) incommensurable in
ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ μήκει. ὡς δὲ ἡ ΑΖ πρὸς τὴν ΖΗ, length with (AG), thus if (some area), equal to the fourth
οὕτως ἐστὶ τὸ ΑΙ πρὸς τὸ ΖΚ· ἀσύμμετρον ἄρα ἐστὶ τὸ part of square on DG, is applied to AG, falling short by
ΑΙ τῷ ΖΚ. καὶ ἐπεὶ αἱ ΑΗ, ΑΓ ῥηταί εἰσι δυνάμει μόνον a square figure, then it divides (AG) into (parts which
σύμμετροι, μέσον ἐστὶ τὸ ΑΚ. πάλιν, ἐπεὶ αἱ ΑΓ, ΔΗ ῥηταί are) incommensurable (in length) [Prop. 10.18]. There-
εἰσι καὶ ἀσύμμετροι μήκει, μέσον ἐστὶ καὶ τὸ ΔΚ. ἐπεὶ οὖν fore, let DG have been cut in half at [point] E. And let
αἱ ΑΗ, ΗΔ δυνάμει μόνον σύμμετροί εἰσιν, ἀσύμμετρος ἄρα (some area), equal to the (square) on EG, have been ap-
ἐστὶν ἡ ΑΗ τῇ ΗΔ μήκει. ὡς δὲ ἡ ΑΗ πρὸς τὴν ΗΔ, οὕτως plied to AG, falling short by a square figure. And let it
ἐστὶ τὸ ΑΚ πρὸς τὸ ΚΔ· ἀσύμμετρον ἄρα ἐστὶ τὸ ΑΚ τῷ be the (rectangle contained) by AF and FG. AF is thus
ΚΔ. incommensurable in length with FG. And as AF (is) to
Συνεστάτω οὖν τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, FG, so AI is to FK [Prop. 6.1]. Thus, AI is incommen-
τῷ δὲ ΖΚ ἴσον ἀφῃρήσθω περὶ τὴν αὐτὴν γωνίαν τὸ ΝΞ· surable with FK [Prop. 10.11]. And since AG and AC
περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα. are rational (straight-lines which are) commensurable in
ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα. square only, AK is a medial (area) [Prop. 10.21]. Again,
ὁμοίως δὴ τοῖς ἐπάνω δείξομεν, ὅτι ἡ ΛΝ δύναται τὸ ΑΒ since AC and DG are rational (straight-lines which are)
χωρίον. λέγω, ὅτι ἡ ΛΝ [ἡ] μετὰ μέσου μέσον τὸ ὅλον incommensurable in length, DK is also a medial (area)
ποιοῦσά ἐστιν. [Prop. 10.21]. Therefore, since AG and GD are com-
᾿Επεὶ γὰρ μέσον ἐδείχθη τὸ ΑΚ καί ἐστιν ἴσον τοῖς ἀπὸ mensurable in square only, AG is thus incommensurable
τῶν ΛΟ, ΟΝ, τὸ ἄρα συγκείμενον ἐκ τῶν ἀπὸ τῶν ΛΟ, ΟΝ in length with GD. And as AG (is) to GD, so AK is to
μέσον ἐστίν. πάλιν, ἐπεὶ μέσον ἐδείχθη τὸ ΔΚ καί ἐστιν KD [Prop. 6.1]. Thus, AK is incommensurable with KD
ἴσον τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ, καὶ τὸ δὶς ὑπὸ τῶν ΛΟ, ΟΝ [Prop. 10.11].
μέσον ἐστίν. καὶ ἐπεὶ ἀσύμμετρον ἐδείχθη τὸ ΑΚ τῷ ΔΚ, Therefore, let the square LM, equal to AI, have been
ἀσύμμετρα [ἄρα] ἐστὶ καὶ τὰ ἀπὸ τῶν ΛΟ, ΟΝ τετράγωνα constructed. And let NO, equal to FK, (and) about the
τῷ δὶς ὑπὸ τῶν ΛΟ, ΟΝ. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ ΑΙ τῷ same angle, have been subtracted (from LM). Thus,
ΖΚ, ἀσύμμετρον ἄρα καὶ τὸ ἀπὸ τῆς ΛΟ τῷ ἀπὸ τῆς ΟΝ· the squares LM and NO are about the same diagonal
αἱ ΛΟ, ΟΝ ἄρα δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τό τε [Prop. 6.26]. Let PR be their (common) diagonal, and
συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον καὶ τὸ δὶς let (the rest of) the figure have been drawn. So, similarly
ὑπ᾿ αὐτῶν μέσον ἔτι τε τὰ ἀπ᾿ αὐτῶν τετράγωνα ἀσύμμετρα to the above, we can show that LN is the square-root of
τῷ δὶς ὑπ᾿ αὐτῶν. ἡ ἄρα ΛΝ ἄλογός ἐστιν ἡ καλουμέμη μετὰ area AB. I say that LN is that (straight-line) which with
μέσου μέσον τὸ ὅλον ποιοῦσα· καὶ δύναται τὸ ΑΒ χωρίον. a medial (area) makes a medial whole.
῾Η ἄρα τὸ χωρίον δυναμένη μετὰ μέσου μέσον τὸ ὅλον For since AK was shown (to be) a medial (area), and
ποιοῦσά ἐστιν· ὅπερ ἔδει δεῖξαι. is equal to the (sum of the) squares on LP and PN, the
sum of the (squares) on LP and PN is medial. Again,
since DK was shown (to be) a medial (area), and is
equal to twice the (rectangle contained) by LP and PN,
twice the (rectangle contained) by LP and PN is also
medial. And since AK was shown (to be) incommensu-
rable with DK, [thus] the (sum of the) squares on LP
and PN is also incommensurable with twice the (rect-
angle contained) by LP and PN. And since AI is in-
commensurable with FK, the (square) on LP (is) thus
also incommensurable with the (square) on PN. Thus,
LP and PN are (straight-lines which are) incommensu-

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