Page 319 - Euclid's Elements of Geometry

ST EW iþ.

q ldþ q ELEMENTS BOOK 10
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† AF and F B have lengths [1 + k/(1 + k ) ]/2 and [1 − k/(1 + k ) ]/2 times that of AB, respectively, where k is defined in the
footnote to Prop. 10.30.
.
Proposition 34
Εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τὸ To find two straight-lines (which are) incommensu-
μὲν συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον, τὸ rable in square, making the sum of the squares on them
δ᾿ ὑπ᾿ αὐτῶν ῥητόν. medial, and the (rectangle contained) by them rational.
∆ 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 ῥητὸν περιέχουσαι τὸ ὑπ᾿ αὐτῶν, ὥστε τὴν ΑΒ (which are) commensurable in square only, be laid out
τῆς ΒΓ μεῖζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ having the (rectangle contained) by them rational, (and)
γεγράφθω ἐπὶ τῆς ΑΒ τὸ ΑΔΒ ἡμικύκλιον, καὶ τετμήσθω such that the square on AB is greater than (the square
ἡ ΒΓ δίχα κατὰ τὸ Ε, καὶ παραβεβλήσθω παρὰ τὴν ΑΒ on) BC by the (square) on (some straight-line) incom-
τῷ ἀπὸ τῆς ΒΕ ἴσον παραλληλόγραμμον ἐλλεῖπον εἴδει τε- mensurable (in length) with (AB) [Prop. 10.31]. And
τραγώνῳ τὸ ὑπὸ τῶν ΑΖΒ· ἀσύμμετρος ἄρα [ἐστὶν] ἡ ΑΖ let the semi-circle ADB have been drawn on AB. And
τῇ ΖΒ μήκει. καὶ ἤχθω ἀπὸ τοῦ Ζ τῇ ΑΒ πρὸς ὀρθὰς ἡ ΖΔ, let BC have been cut in half at E. And let a (rectangu-
καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΒ. lar) parallelogram equal to the (square) on BE, (and)
᾿Επεὶ ἀσύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΒ, ἀσύμμετρον ἄρα falling short by a square figure, have been applied to
ἐστὶ καὶ τὸ ὑπὸ τῶν ΒΑ, ΑΖ τῷ ὑπὸ τῶν ΑΒ, ΒΖ. ἴσον δὲ AB, (and let it be) the (rectangle contained by) AFB
τὸ μὲν ὑπὸ τῶν ΒΑ, ΑΖ τῷ ἀπὸ τῆς ΑΔ, τὸ δὲ ὑπὸ τῶν [Prop. 6.28]. Thus, AF [is] incommensurable in length
ΑΒ, ΒΖ τῷ ἀπὸ τῆς ΔΒ· ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ with FB [Prop. 10.18]. And let FD have been drawn
τῆς ΑΔ τῷ ἀπὸ τῆς ΔΒ. καὶ ἐπεὶ μέσον ἐστὶ τὸ ἀπὸ τῆς from F at right-angles to AB. And let AD and DB have
ΑΒ, μέσον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, been joined.
ΔΒ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΒΓ τῆς ΔΖ, διπλάσιον ἄρα καὶ Since AF is incommensurable (in length) with FB,
τὸ ὑπὸ τῶν ΑΒ, ΒΓ τοῦ ὑπὸ τῶν ΑΒ, ΖΔ. ῥητὸν δὲ τὸ ὑπὸ the (rectangle contained) by BA and AF is thus also
τῶν ΑΒ, ΒΓ· ῥητὸν ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ. τὸ δὲ ὑπὸ incommensurable with the (rectangle contained) by AB
τῶν ΑΒ, ΖΔ ἴσον τῷ ὑπὸ τῶν ΑΔ, ΔΒ· ὥστε καὶ τὸ ὑπὸ and BF [Prop. 10.11]. And the (rectangle contained) by
τῶν ΑΔ, ΔΒ ῥητόν ἐστιν. BA and AF (is) equal to the (square) on AD, and the
Εὕρηνται ἄρα δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΔ, (rectangle contained) by AB and BF to the (square) on
ΔΒ ποιοῦσαι τὸ [μὲν] συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν τε- DB [Prop. 10.32 lem.]. Thus, the (square) on AD is also
τραγώνων μέσον, τὸ δ᾿ ὑπ᾿ αὐτῶν ῥητόν· ὅπερ ἔδει δεῖξαι. incommensurable with the (square) on DB. And since
the (square) on AB is medial, the sum of the (squares)
on AD and DB (is) thus also medial [Props. 3.31, 1.47].
And since BC is double DF [see previous proposition],
the (rectangle contained) by AB and BC (is) thus also
double the (rectangle contained) by AB and FD. And
the (rectangle contained) by AB and BC (is) rational.
Thus, the (rectangle contained) by AB and FD (is) also
rational [Prop. 10.6, Def. 10.4]. And the (rectangle con-
tained) by AB and FD (is) equal to the (rectangle con-
tained) by AD and DB [Prop. 10.32 lem.]. And hence
the (rectangle contained) by AD and DB is rational.
Thus, two straight-lines, AD and DB, (which are) in-
commensurable in square, have been found, making the
sum of the squares on them medial, and the (rectangle

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