Page 43 - Euclid's Elements of Geometry
P. 43

ST	EW      aþ.











            πληρώματί ἐστιν ἴσον.                               the remaining complement KD. ELEMENTS BOOK 1
                       Α         Θ                      ∆                 A         H                      D


                                Κ                                                   K
                   Ε                                   Ζ               E                                  F


                                   mdþ



                Β         Η                      Γ                  B        G                       C
               Παντὸς ἄρα παραλληλογράμμου χωρίου τῶν περὶ τὴν     Thus, for any parallelogramic figure, the comple-
            διάμετρον παραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλή- ments of the parallelograms about the diagonal are equal
            λοις ἐστίν· ὅπερ ἔδει δεῖξαι.                       to one another. (Which is) the very thing it was required
                                                                to show.

                                                                                 Proposition 44
                                      .
               Παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι τριγώνῳ ἴσον πα-  To apply a parallelogram equal to a given triangle to
            ραλληλόγραμμον παραβαλεῖν ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμ- a given straight-line in a given rectilinear angle.
            μῳ.


                                               ∆                                                  D
                                 Γ                                                   C


                             Ζ     Ε            Κ                               F      E            K







                        Η                      Μ                            G                      M
                                  Β                                                  B


                         Θ     Α             Λ                              H      A            L
               ῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν τρίγωνον  Let AB be the given straight-line, C the given trian-
            τὸ Γ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ· δεῖ δὴ παρὰ gle, and D the given rectilinear angle. So it is required to
            τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι τριγώνῳ τῷ Γ  apply a parallelogram equal to the given triangle C to the
            ἴσον παραλληλόγραμμον παραβαλεῖν ἐν ἴσῃ τῇ Δ γωνίᾳ.  given straight-line AB in an angle equal to (angle) D.
               Συνεστάτω τῷ Γ τριγώνῳ ἴσον παραλληλόγραμμον τὸ     Let the parallelogram BEFG, equal to the triangle C,
            ΒΕΖΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ, ἥ ἐστιν ἴση τῇ Δ· καὶ κείσθω have been constructed in the angle EBG, which is equal
            ὥστε ἐπ᾿ εὐθείας εἶναι τὴν ΒΕ τῇ ΑΒ, καὶ διήχθω ἡ ΖΗ to D [Prop. 1.42]. And let it have been placed so that
                                                                                      †
            ἐπὶ τὸ Θ, καὶ διὰ τοῦ Α ὁποτέρᾳ τῶν ΒΗ, ΕΖ παράλληλος BE is straight-on to AB. And let FG have been drawn
            ἤχθω ἡ ΑΘ, καὶ ἐπεζεύχθω ἡ ΘΒ. καὶ ἐπεὶ εἰς παραλλήλους through to H, and let AH have been drawn through A
            τὰς ΑΘ, ΕΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ, αἱ ἄρα ὑπὸ ΑΘΖ, ΘΖΕ parallel to either of BG or EF [Prop. 1.31], and let HB
            γωνίαι δυσὶν ὀρθαῖς εἰσιν ἴσαι. αἱ ἄρα ὑπὸ ΒΘΗ, ΗΖΕ  have been joined. And since the straight-line HF falls
            δύο ὀρθῶν ἐλάσσονές εἰσιν· αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο  across the parallels AH and EF, the (sum of the) an-
            ὀρθῶν εἰς ἄπειρον ἐκβαλλόμεναι συμπίπτουσιν· αἱ ΘΒ, ΖΕ gles AHF and HFE is thus equal to two right-angles


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