Page 294 - Euclid's Elements of Geometry

ST EW iþ.

ELEMENTS BOOK 10

(straight-line is) larger than (the square on) the lesser. †
Γ C
∆ D

Α Β A B
῎Εστωσαν αἱ δοθεῖσαι δύο ἄνισοι εὐθεῖαι αἱ ΑΒ, Γ, ὧν Let AB and C be the two given unequal straight-lines,
μείζων ἔστω ἡ ΑΒ· δεῖ δὴ εὑρεῖν, τίνι μεῖζον δύναται ἡ ΑΒ and let AB be the greater of them. So it is required to
τῆς Γ. find by (the square on) which (straight-line) the square
Γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ εἰς αὐτὸ on AB (is) greater than (the square on) C.
ἐνηρμόσθω τῇ Γ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΒ. φανερὸν Let the semi-circle ADB have been described on AB.
δή, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΑΔΒ γωνία, καὶ ὅτι ἡ ΑΒ τῆς And let AD, equal to C, have been inserted into it
ΑΔ, τουτέστι τῆς Γ, μεῖζον δύναται τῇ ΔΒ. [Prop. 4.1]. And let DB have been joined. So (it is) clear
῾Ομοίως δὲ καὶ δύο δοθεισῶν εὐθειῶν ἡ δυναμένη αὐτὰς that the angle ADB is a right-angle [Prop. 3.31], and
εὑρίσκεται οὕτως. that the square on AB (is) greater than (the square on)
῎Εστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ, καὶ δέον AD—that is to say, (the square on) C—by (the square
ἔστω εὑρεῖν τὴν δυναμένην αὐτάς. κείσθωσαν γάρ, ὥστε on) DB [Prop. 1.47].
ὀρθὴν γωνίαν περιέχειν τὴν ὑπὸ ΑΔ, ΔΒ, καὶ ἐπεζεύχθω And, similarly, the square-root of (the sum of the
ἡ ΑΒ· φανερὸν πάλιν, ὅτι ἡ τὰς ΑΔ, ΔΒ δυναμένη ἐστὶν ἡ squares on) two given straight-lines is also found likeso.
ΑΒ· ὅπερ ἔδει δεῖξαι. Let AD and DB be the two given straight-lines. And
let it be necessary to find the square-root of (the sum
of the squares on) them. For let them have been laid
idþ been joined. (It is) again clear that AB is the square-root
down such as to encompass a right-angle—(namely), that
(angle encompassed) by AD and DB. And let AB have

of (the sum of the squares on) AD and DB [Prop. 1.47].
(Which is) the very thing it was required to show.

† That is, if α and β are the lengths of two given straight-lines, with α being greater than β, to find a straight-line of length γ such that
2
2
2
2
2
2
α = β + γ . Similarly, we can also find γ such that γ = α + β .
Proposition 14
.
᾿Εὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, δύνηται δὲ ἡ If four straight-lines are proportional, and the square
πρώτη τῆς δευτέρας μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ on the first is greater than (the square on) the sec-
[μήκει], καὶ ἡ τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ ond by the (square) on (some straight-line) commen-
συμμέτρου ἑαυτῇ [μήκει]. καὶ ἐὰν ἡ πρώτη τῆς δευτέρας surable [in length] with the first, then the square on
μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ [μήκει], καὶ ἡ the third will also be greater than (the square on) the
τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου fourth by the (square) on (some straight-line) commen-
ἑαυτῇ [μήκει]. surable [in length] with the third. And if the square on
῎Εστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, Δ, the first is greater than (the square on) the second by
ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, καὶ ἡ Α μὲν the (square) on (some straight-line) incommensurable
τῆς Β μεῖζον δυνάσθω τῷ ἀπὸ τῆς Ε, ἡ δὲ Γ τῆς Δ μεῖζον [in length] with the first, then the square on the third
δυνάσθω τῷ ἀπὸ τῆς Ζ· λέγω, ὅτι, εἴτε σύμμετρός ἐστιν will also be greater than (the square on) the fourth by
ἡ Α τῇ Ε, σύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός the (square) on (some straight-line) incommensurable
ἐστιν ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ὁ Γ τῇ Ζ. [in length] with the third.
Let A, B, C, D be four proportional straight-lines,
(such that) as A (is) to B, so C (is) to D. And let the
square on A be greater than (the square on) B by the

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