Page 320 - Euclid's Elements of Geometry

ST EW iþ.

leþ contained) by them rational. (Which is) the very thing it
ELEMENTS BOOK 10

†
was required to show.

q q
† AD and DB have lengths [(1 + k ) + k]/[2 (1 + k )] and [(1 + k ) − k]/[2 (1 + k )] times that of AB, respectively, where k is
2 1/2
2
2
2 1/2
defined in the footnote to Prop. 10.29.
.
Proposition 35
Εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τό To find two straight-lines (which are) incommensu-
τε συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον καὶ rable in square, making the sum of the squares on them
τὸ ὑπ᾿ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ medial, and the (rectangle contained) by them medial,
τῶν ἀπ᾿ αὐτῶν τετραγώνῳ. and, moreover, incommensurable with the sum of the
squares on them.
∆ D

Α Ζ Β Ε Γ A F B E C
᾿Εκκείσθωσαν δύο μέσαι δυνάμει μόνον σύμμετροι αἱ Let the two medial (straight-lines) AB and BC,
ΑΒ, ΒΓ μέσον περιέχουσαι, ὥστε τὴν ΑΒ τῆς ΒΓ μεῖζον (which are) commensurable in square only, be laid out
δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ γεγράφθω ἐπὶ τῆς containing a medial (area), such that the square on AB
ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ τὰ λοιπὰ γεγονέτω τοῖς ἐπάνω is greater than (the square on) BC by the (square) on
ὁμοίως. (some straight-line) incommensurable (in length) with
Καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΒ μήκει, ἀσύμμετρ- (AB) [Prop. 10.32]. And let the semi-circle ADB have
ός ἐστι καὶ ἡ ΑΔ τῇ ΔΒ δυνάμει. καὶ ἐπεὶ μέσον ἐστὶ τὸ been drawn on AB. And let the remainder (of the figure)
ἀπὸ τῆς ΑΒ, μέσον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ be generated similarly to the above (proposition).
τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΖ, ΖΒ ἴσον ἐστὶ τῷ ἀφ᾿ And since AF is incommensurable in length with FB
ἑκατέρας τῶν ΒΕ, ΔΖ, ἴση ἄρα ἐστὶν ἡ ΒΕ τῇ ΔΖ· διπλῆ [Prop. 10.18], AD is also incommensurable in square
ἄρα ἡ ΒΓ τῆς ΖΔ· ὥστε καὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ διπλάσιόν with DB [Prop. 10.11]. And since the (square) on AB
ἐστι τοῦ ὑπὸ τῶν ΑΒ, ΖΔ. μέσον δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ· is medial, the sum of the (squares) on AD and DB (is)
μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ. καί ἐστιν ἴσον τῷ ὑπὸ thus also medial [Props. 3.31, 1.47]. And since the (rect-
τῶν ΑΔ, ΔΒ· μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ angle contained) by AF and FB is equal to the (square)
ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, σύμμετρος δὲ ἡ ΓΒ on each of BE and DF, BE is thus equal to DF. Thus,
τῇ ΒΕ, ἀσύμμετρος ἄρα καὶ ἡ ΑΒ τῇ ΒΕ μήκει· ὥστε καὶ τὸ BC (is) double FD. And hence the (rectangle contained)
ἀπὸ τῆς ΑΒ τῷ ὑπὸ τῶν ΑΒ, ΒΕ ἀσύμμετρόν ἐστιν. ἀλλὰ by AB and BC is double the (rectangle) contained by
τῷ μὲν ἀπὸ τῆς ΑΒ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ, τῷ δὲ AB and FD. And the (rectangle contained) by AB and
ὑπὸ τῶν ΑΒ, ΒΕ ἴσον ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ, τουτέστι BC (is) medial. Thus, the (rectangle contained) by AB
τὸ ὑπὸ τῶν ΑΔ, ΔΒ· ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον and FD (is) also medial. And it is equal to the (rectan-
ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ τῷ ὑπὸ τῶν ΑΔ, ΔΒ. gle contained) by AD and DB [Prop. 10.32 lem.]. Thus,
Εὕρηνται ἄρα δύο εὐθεῖαι αἱ ΑΔ, ΔΒ δυνάμει ἀσύμμετροι the (rectangle contained) by AD and DB (is) also me-
ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν μέσον καὶ dial. And since AB is incommensurable in length with
τὸ ὑπ᾿ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ BC, and CB (is) commensurable (in length) with BE,
τῶν ἀπ᾿ αὐτῶν τετραγώνων· ὅπερ ἔδει δεῖξαι. AB (is) thus also incommensurable in length with BE
[Prop. 10.13]. And hence the (square) on AB is also
incommensurable with the (rectangle contained) by AB
and BE [Prop. 10.11]. But the (sum of the squares) on
AD and DB is equal to the (square) on AB [Prop. 1.47].
And the (rectangle contained) by AB and FD—that is to
say, the (rectangle contained) by AD and DB—is equal
to the (rectangle contained) by AB and BE. Thus, the

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