Page 16 - Euclid's Elements of Geometry
P. 16
ST EW aþ.
ELEMENTS BOOK 1
ἡ ΑΔ βάσει τῇ ΒΔ ἴση ἐστίν. the two (straight-lines) AC, CD are equal to the two
(straight-lines) BC, CD, respectively. And the angle
ACD is equal to the angle BCD. Thus, the base AD
is equal to the base BD [Prop. 1.4].
Γ C
iaþ ∆ D
Α Β A B
῾Η ἄρα δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ δίχα τέτμηται Thus, the given finite straight-line AB has been cut
κατὰ τὸ Δ· ὅπερ ἔδει ποιῆσαι. in half at (point) D. (Which is) the very thing it was
required to do.
Proposition 11
.
Τῇ δοθείσῃ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου To draw a straight-line at right-angles to a given
πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν. straight-line from a given point on it.
Ζ F
Α Β A B
∆ Γ Ε D C E
῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ τὸ δὲ δοθὲν σημεῖον Let AB be the given straight-line, and C the given
ἐπ᾿ αὐτῆς τὸ Γ· δεῖ δὴ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ point on it. So it is required to draw a straight-line from
πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν. the point C at right-angles to the straight-line AB.
Εἰλήφθω ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ, καὶ κείσθω Let the point D be have been taken at random on AC,
τῇ ΓΔ ἴση ἡ ΓΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον and let CE be made equal to CD [Prop. 1.3], and let the
ἰσόπλευρον τὸ ΖΔΕ, καὶ ἐπεζεύχθω ἡ ΖΓ· λέγω, ὅτι τῇ equilateral triangle FDE have been constructed on DE
δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου [Prop. 1.1], and let FC have been joined. I say that the
τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ. straight-line FC has been drawn at right-angles to the
᾿Επεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ, κοινὴ δὲ ἡ ΓΖ, δύο given straight-line AB from the given point C on it.
δὴ αἱ ΔΓ, ΓΖ δυσὶ ταῖς ΕΓ, ΓΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· For since DC is equal to CE, and CF is common,
καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ ἴση ἐστίν· γωνία ἄρα ἡ ὑπὸ the two (straight-lines) DC, CF are equal to the two
ΔΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ ἴση ἐστίν· καί εἰσιν ἐφεξῆς. ὅταν (straight-lines), EC, CF, respectively. And the base DF
δὲ εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας is equal to the base FE. Thus, the angle DCF is equal
ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν· ὀρθὴ to the angle ECF [Prop. 1.8], and they are adjacent.
ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΔΓΖ, ΖΓΕ. But when a straight-line stood on a(nother) straight-line
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