Page 41 - Euclid's Elements of Geometry

ST EW aþ.

ELEMENTS BOOK 1

ἐλάσσονι· ὅπερ ἐστὶν ἀδύνατον· οὐκ ἄρα παράλληλος ἡ ΑΖ to [triangle] DCE. Thus, [triangle] DCE is also equal to
τῇ ΒΕ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾿ ἄλλη τις πλὴν τῆς ΑΔ· triangle FCE, the greater to the lesser. The very thing is
maþ on the same side, are also between the same parallels.
ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος. impossible. Thus, AF is not parallel to BE. Similarly, we
Τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ can show that neither (is) any other (straight-line) than
αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν· ὅπερ ἔδει AD. Thus, AD is parallel to BE.
δεῖξαι. Thus, equal triangles which are on equal bases, and

(Which is) the very thing it was required to show.

† This whole proposition is regarded by Heiberg as a relatively early interpolation to the original text.

Proposition 41
.
᾿Εὰν παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν If a parallelogram has the same base as a triangle, and
αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστί is between the same parallels, then the parallelogram is
τὸ παραλληλόγραμμον τοῦ τριγώνου. double (the area) of the triangle.
Α ∆ Ε A D E

Β Γ B C
Παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ τριγώνῳ τῷ ΕΒΓ For let parallelogram ABCD have the same base BC
βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ καὶ ἐν ταῖς αὐταῖς πα- as triangle EBC, and let it be between the same parallels,
ραλλήλοις ἔστω ταῖς ΒΓ, ΑΕ· λέγω, ὅτι διπλάσιόν ἐστι τὸ BC and AE. I say that parallelogram ABCD is double
ΑΒΓΔ παραλληλόγραμμον τοῦ ΒΕΓ τριγώνου. (the area) of triangle BEC.
᾿Επεζεύχθω γὰρ ἡ ΑΓ. ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον For let AC have been joined. So triangle ABC is equal
τῷ ΕΒΓ τριγώνῳ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν to triangle EBC. For it is on the same base, BC, as
αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΕ. (EBC), and between the same parallels, BC and AE
mbþ
ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον διπλάσιόν ἐστι τοῦ ΑΒΓ [Prop. 1.37]. But, parallelogram ABCD is double (the
τριγώνου· ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει· ὥστε area) of triangle ABC. For the diagonal AC cuts the for-
τὸ ΑΒΓΔ παραλληλόγραμμον καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ mer in half [Prop. 1.34]. So parallelogram ABCD is also
διπλάσιον. double (the area) of triangle EBC.
᾿Εὰν ἄρα παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν Thus, if a parallelogram has the same base as a trian-
αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστί τὸ gle, and is between the same parallels, then the parallel-
παραλληλόγραμμον τοῦ τριγώνου· ὅπερ ἔδει δεῖξαι. ogram is double (the area) of the triangle. (Which is) the
very thing it was required to show.

.
Proposition 42
Τῷ δοθέντι τριγώνῳ ἴσον παραλληλόγραμμον συστή- To construct a parallelogram equal to a given triangle
σασθαι ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ. in a given rectilinear angle.
῎Εστω τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ, ἡ δὲ δοθεῖσα Let ABC be the given triangle, and D the given recti-
γωνία εὐθύγραμμος ἡ Δ· δεῖ δὴ τῷ ΑΒΓ τριγώνῳ ἴσον πα- linear angle. So it is required to construct a parallelogram
ραλληλόγραμμον συστήσασθαι ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ. equal to triangle ABC in the rectilinear angle D.

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