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Euclid · Elements §1.prop.39-1.prop.41

Equal Triangles in Parallels and Parallelograms as Double

Passage 21 of 316 · Greek

Summary

These propositions prove that equal triangles on the same (or equal) bases and on the same side are in the same parallels (Props. 39 and 40). They also demonstrate that a parallelogram is double of a triangle when they share the same base and are in the same parallels (Prop. 41).

§1.prop.39τὰ ἴσα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν.
Equal triangles which are on the same base and on the same side are also in the same parallels.
ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ· λέγω, ὅτι καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν.
Let ABC, DBC be equal triangles on the same base BC and on the same side; I say that they are also in the same parallels.
ἐπεζεύχθω γὰρ ἡ ΑΔ· λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ. εἰ γὰρ μή, ἤχθω διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ παράλληλος ἡ ΑΕ, καὶ ἐπεζεύχθω ἡ ΕΓ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΕΒΓ τριγώνῳ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις.
For let AD be joined; I say that AD is parallel to BC. For if not, let AE be drawn through the point A parallel to the straight line BC, and let EC be joined. Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels.
ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον· καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ τὸ μεῖζον τῷ ἐλάσσονι· ὅπερ ἐστὶν ἀδύνατον·
But ABC is equal to DBC; therefore DBC is also equal to EBC, the greater to the less; which is impossible.
οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ. ὁμοίως δὴ δείξομεν, ὅτι οὐδʼ ἄλλη τις πλὴν τῆς ΑΔ· ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος.
Therefore AE is not parallel to BC. In the same way we can prove that no other straight line is parallel except AD; therefore AD is parallel to BC.
τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
Therefore equal triangles which are on the same base and on the same side are also in the same parallels; which was to be proved.
§1.prop.40τὰ ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν.
Equal triangles which are on equal bases and on the same side are also in the same parallels.
ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΓΔΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ, ΓΕ καὶ ἐπὶ τὰ αὐτὰ μέρη. λέγω, ὅτι καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν.
Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side; I say that they are also in the same parallels.
ἐπεζεύχθω γὰρ ἡ ΑΔ· λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ. εἰ γὰρ μή, ἤχθω διὰ τοῦ Α τῇ ΒΕ παράλληλος ἡ ΑΖ, καὶ ἐπεζεύχθω ἡ ΖΕ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΖΓΕ τριγώνῳ· ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΓΕ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΕ, ΑΖ. ἀλλὰ τὸ ΑΒΓ τρίγωνον ἴσον ἐστὶ τῷ ΔΓΕ· καὶ τὸ ΔΓΕ ἄρα ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ τὸ μεῖζον τῷ ἐλάσσονι· ὅπερ ἐστὶν ἀδύνατον·
For let AD be joined; I say that AD is parallel to BE. For if not, let AZ be drawn through A parallel to BE, and let ZE be joined. Therefore the triangle ABC is equal to the triangle ZCE; for they are on equal bases BC, CE and in the same parallels BE, AZ. But the triangle ABC is equal to DCE; therefore DCE is also equal to the triangle ZCE, the greater to the less; which is impossible.
οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ. ὁμοίως δὴ δείξομεν, ὅτι οὐδʼ ἄλλη τις πλὴν τῆς ΑΔ· ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος.
Therefore AZ is not parallel to BE. In the same way we can prove that no other straight line is parallel except AD; therefore AD is parallel to BE.
τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
Therefore equal triangles which are on equal bases and on the same side are also in the same parallels; which was to be proved.
§1.prop.41ἐὰν παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστι τὸ παραλληλόγραμμον τοῦ τριγώνου.
If a parallelogram has the same base with a triangle and is in the same parallels, the parallelogram is double of the triangle.
παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ τριγώνῳ τῷ ΕΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ταῖς ΒΓ, ΑΕ· λέγω, ὅτι διπλάσιόν ἐστι τὸ ΑΒΓΔ παραλληλόγραμμον τοῦ ΒΕΓ τριγώνου.
For let a parallelogram ABCD have the same base BC with a triangle EBC, and let it be in the same parallels BC, AE; I say that the parallelogram ABCD is double of the triangle BEC.
ἐπεζεύχθω γὰρ ἡ ΑΓ. ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΕΒΓ τριγώνῳ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΕ. ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου· ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει· ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον.
For let AC be joined. Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels BC, AE. But the parallelogram ABCD is double of the triangle ABC; for the diameter AC bisects it; so that the parallelogram ABCD is also double of the triangle EBC.
ἐὰν ἄρα παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστι τὸ παραλληλόγραμμον τοῦ τριγώνου· ὅπερ ἔδει δεῖξαι.
Therefore, if a parallelogram has the same base with a triangle and is in the same parallels, the parallelogram is double of the triangle; which was to be proved.

Notes

  1. §1.prop.39εἰ γὰρ μή — An idiomatic ellipsis meaning "for if not." Here it stands for "for if AD is not parallel to BC" (εἰ γὰρ μὴ ἔστι παράλληλος ἡ ΑΔ τῇ ΒΓ).
  2. §1.prop.39τὸ μεῖζον τῷ ἐλάσσονι — Connected with the preceding ἴσον ἐστὶ, it explains the specific nature of the contradiction as an appositional supplement: "[namely] the greater [equal] to the less," contrastingly emphasizing the absurdity of the previous statement.
  3. §1.prop.41βάσιν τε ἔχῃ τὴν αὐτὴν — Used with the dative τριγώνῳ to mean "has the same (τὴν αὐτὴν) base as the triangle." τὴν αὐτὴν corresponds with the dative to express identity.

Cite this passage

Euclid, Elements §1.prop.39-1.prop.41. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:1.prop.39-1.prop.41

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