Page 332 - Euclid's Elements of Geometry

ST EW iþ.

C
B
Α ∆ Γ Β A D ELEMENTS BOOK 10

Ε Ζ E F

Μ Λ M L
Θ Η H G

Ν Κ N K
῎Εστω [δύο μέσα δυναμένη] ἡ ΑΒ διῃρημένη κατὰ τὸ Let AB be [the square-root of (the sum of) two me-
Γ, ὥστε τὰς ΑΓ, ΓΒ δυνάμει ἀσυμμέτρους εἶναι ποιούσας dial (areas)] which has been divided at C, such that AC
τό τε συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μέσον καὶ τὸ and CB are incommensurable in square, making the sum
ὑπὸ τῶν ΑΓ, ΓΒ μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ of the (squares) on AC and CB medial, and the (rect-
ἐκ τῶν ἀπ᾿ αὐτῶν. λέγω, ὅτι ἡ ΑΒ κατ᾿ ἄλλο σημεῖον οὐ angle contained) by AC and CB medial, and, moreover,
διαιρεῖται ποιοῦσα τὰ προκείμενα. incommensurable with the sum of the (squares) on (AC
Εἰ γὰρ δυνατόν, διῃρήσθω κατὰ τὸ Δ, ὥστε πάλιν δη- and CB) [Prop. 10.41]. I say that AB cannot be divided
λονότι τὴν ΑΓ τῇ ΔΒ μὴ εἶναι τὴν αὐτήν, ἀλλὰ μείζονα at another point fulfilling the prescribed (conditions).
καθ᾿ ὑπόθεσιν τὴν ΑΓ, καὶ ἐκκείσθω ῥητὴ ἡ ΕΖ, καὶ παρα- For, if possible, let it have been divided at D, such that
βεβλήσθω παρὰ τὴν ΕΖ τοῖς μὲν ἀπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ AC is again manifestly not the same as DB, but AC (is),
ΕΗ, τῷ δὲ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἴσον τὸ ΘΚ· ὅλον ἄρα τὸ by hypothesis, greater. And let the rational (straight-line)
ΕΚ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. πάλιν δὴ παρα- EF be laid down. And let EG, equal to (the sum of) the
βεβλήσθω παρὰ τὴν ΕΖ τοῖς ἀπὸ τῶν ΑΔ, ΔΒ ἴσον τὸ ΕΛ· (squares) on AC and CB, and HK, equal to twice the
λοιπὸν ἄρα τὸ δὶς ὑπὸ τῶν ΑΔ, ΔΒ λοιπῷ τῷ ΜΚ ἴσον (rectangle contained) by AC and CB, have been applied
ἐστίν. καὶ ἐπεὶ μέσον ὑπόκειται τὸ συγκείμενον ἐκ τῶν ἀπὸ to EF. Thus, the whole of EK is equal to the square on
τῶν ΑΓ, ΓΒ, μέσον ἄρα ἐστὶ καὶ τὸ ΕΗ. καὶ παρὰ ῥητὴν τὴν AB [Prop. 2.4]. So, again, let EL, equal to (the sum of)
ΕΖ παράκειται· ῥητὴ ἄρα ἐστὶν ἡ ΘΕ καὶ ἀσύμμετρος τῇ ΕΖ the (squares) on AD and DB, have been applied to EF.
μήκει. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΝ ῥητή ἐστι καὶ ἀσύμμετρος Thus, the remainder—twice the (rectangle contained) by
τῇ ΕΖ μήκει. καὶ ἐπεὶ ἀσύμμετρόν ἐστι τὸ συγκείμενον AD and DB—is equal to the remainder, MK. And since
ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ, καὶ τὸ the sum of the (squares) on AC and CB was assumed
ΕΗ ἄρα τῷ ΗΝ ἀσύμμετρόν ἐστιν· ὥστε καὶ ἡ ΕΘ τῇ ΘΝ (to be) medial, EG is also medial. And it is applied to
ἀσύμμετρός ἐστιν. καί εἰσι ῥηταί· αἱ ΕΘ, ΘΝ ἄρα ῥηταί εἰσι the rational (straight-line) EF. HE is thus rational, and
δυνάμει μόνον σύμμετροι· ἡ ΕΝ ἄρα ἐκ δύο ὀνομάτων ἐστὶ incommensurable in length with EF [Prop. 10.22]. So,
διῃρημένη κατὰ τὸ Θ. ὁμοίως δὴ δείξομεν, ὅτι καὶ κατὰ τὸ for the same (reasons), HN is also rational, and incom-
Μ διῄρηται. καὶ οὐκ ἔστιν ἡ ΕΘ τῇ ΜΝ ἡ αὐτή· ἡ ἄρα ἐκ δύο mensurable in length with EF. And since the sum of
ὀνομάτων κατ᾿ ἄλλο καὶ ἄλλο σημεῖον διῄρηται· ὅπερ ἐστίν the (squares) on AC and CB is incommensurable with
ἄτοπον. οὐκ ἄρα ἡ δύο μέσα δυναμένη κατ᾿ ἀλλο καὶ ἄλλο twice the (rectangle contained) by AC and CB, EG is
σημεῖον διαιρεῖται· καθ᾿ ἓν ἄρα μόνον [σημεῖον] διαιρεῖται. thus also incommensurable with GN. Hence, EH is also
incommensurable with HN [Props. 6.1, 10.11]. And
they are (both) rational (straight-lines). Thus, EH and
HN are rational (straight-lines which are) commensu-
rable in square only. Thus, EN is a binomial (straight-
line) which has been divided (into its component terms)
at H [Prop. 10.36]. So, similarly, we can show that it has
also been (so) divided at M. And EH is not the same as
MN. Thus, a binomial (straight-line) has been divided
(into its component terms) at different points. The very
thing is absurd [Prop. 10.42]. Thus, the square-root of
(the sum of) two medial (areas) cannot be divided (into

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