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Euclid · Elements §11.prop.35#1

Angles Formed by Elevated Lines and Projected Lines

Passage 264 of 316 · Greek

Summary

It is proved that if two elevated straight lines set up from the vertices of two equal plane angles contain equal angles with the original sides, then the straight lines joining the vertices with the feet of the perpendiculars from the elevated lines contain equal angles with the elevated lines. Corresponding line segments are also shown to be equal.

§11.prop.35#1ἐὰν ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι, ἐπὶ δὲ τῶν κορυφῶν αὐτῶν μετέωροι εὐθεῖαι ἐπισταθῶσιν ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, ἐπὶ δὲ τῶν μετεώρων ληφθῇ τυχόντα σημεῖα, καὶ ἀπʼ αὐτῶν ἐπὶ τὰ ἐπίπεδα, ἐν οἷς εἰσιν αἱ ἐξ ἀρχῆς γωνίαι, κάθετοι ἀχθῶσιν, ἀπὸ δὲ τῶν γενομένων σημείων ἐν τοῖς ἐπιπέδοις ἐπὶ τὰς ἐξ ἀρχῆς γωνίας ἐπιζευχθῶσιν εὐθεῖαι, ἴσας γωνίας περιέξουσι μετὰ τῶν μετεώρων.
If there be two equal plane angles, and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines, each to each, and on the elevated straight lines random points be taken, and from them perpendiculars be drawn to the planes in which the original angles are, and from the points produced in the planes straight lines be joined to the vertices of the original angles, they will contain equal angles with the elevated straight lines.
ἔστωσαν δύο γωνίαι εὐθύγραμμοι ἴσαι αἱ ὑπὸ ΒΑΓ, ΕΔΖ, ἀπὸ δὲ τῶν Α, Δ σημείων μετέωροι εὐθεῖαι ἐφεστάτωσαν αἱ ΑΗ, ΔΜ ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, τὴν μὲν ὑπὸ ΜΔΕ τῇ ὑπὸ ΗΑΒ, τὴν δὲ ὑπὸ ΜΔΖ τῇ ὑπὸ ΗΑΓ, καὶ εἰλήφθω ἐπὶ τῶν ΑΗ, ΔΜ τυχόντα σημεῖα τὰ Η, Μ, καὶ ἤχθωσαν ἀπὸ τῶν Η, Μ σημείων ἐπὶ τὰ διὰ τῶν ΒΑΓ, ΕΔΖ ἐπίπεδα κάθετοι αἱ ΗΛ, ΜΝ, καὶ συμβαλλέτωσαν τοῖς ἐπιπέδοις κατὰ τὰ Ν, Λ, καὶ ἐπεζεύχθωσαν αἱ ΛΑ, ΝΔ· λέγω, ὅτι ἴση ἐστὶν ἡ ὑπὸ ΗΑΛ γωνία τῇ ὑπὸ ΜΔΝ γωνίᾳ.
Let there be two equal rectilineal angles BAG, EDZ, and from the points A, D let elevated straight lines AH, DM be set up containing equal angles with the original straight lines, each to each, namely MDE to HAB, and MDZ to HAG, and let random points H, M be taken on AH, DM, and from the points H, M let perpendiculars HL, MN be drawn to the planes through BAG, EDZ, and let them meet the planes at N, L, and let LA, ND be joined; I say that the angle HAL is equal to the angle MDN.
κείσθω τῇ ΔΜ ἴση ἡ ΑΘ, καὶ ἤχθω διὰ τοῦ Θ σημείου τῇ ΗΛ παράλληλος ἡ ΘΚ. ἡ δὲ ΗΛ κάθετός ἐστιν ἐπὶ τὸ διὰ τῶν ΒΑΓ ἐπίπεδον· καὶ ἡ ΘΚ ἄρα κάθετός ἐστιν ἐπὶ τὸ διὰ τῶν ΒΑΓ ἐπίπεδον.
Let AQ be laid down equal to DM, and through the point Q let QK be drawn parallel to HL. But HL is perpendicular to the plane through BAG; therefore QK is also perpendicular to the plane through BAG.
ἤχθωσαν ἀπὸ τῶν Κ, Ν σημείων ἐπὶ τὰς ΑΒ, ΑΓ, ΔΖ, ΔΕ εὐθείας κάθετοι αἱ ΚΓ, ΝΖ, ΚΒ, ΝΕ, καὶ ἐπεζεύχθωσαν αἱ ΘΓ, ΓΒ, ΜΖ, ΖΕ. ἐπεὶ τὸ ἀπὸ τῆς ΘΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΑ, τῷ δὲ ἀπὸ τῆς ΚΑ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΚΓ, ΓΑ, καὶ τὸ ἀπὸ τῆς ΘΑ ἄρα ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΓ, ΓΑ. τοῖς δὲ ἀπὸ τῶν ΘΚ, ΚΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΘΓ·
From the points K, N let perpendiculars KG, NZ, KB, NE be drawn to the straight lines AB, AG, DZ, DE, and let QG, GB, MZ, ZE be joined. Since the square on QA is equal to the squares on QK, KA, and the squares on KG, GA are equal to the square on KA, the square on QA is also equal to the squares on QK, KG, GA.
τὸ ἄρα ἀπὸ τῆς ΘΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΓ, ΓΑ. ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΘΓΑ γωνία.
But the square on QG is equal to the squares on QK, KG; therefore the square on QA is equal to the squares on QG, GA. Therefore the angle QGA is right.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΔΖΜ γωνία ὀρθή ἐστιν.
For the same reasons indeed the angle DZM is also right.
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΓΘ γωνία τῇ ὑπὸ ΔΖΜ. ἔστι δὲ καὶ ἡ ὑπὸ ΘΑΓ τῇ ὑπὸ ΜΔΖ ἴση.
Therefore the angle AGQ is equal to the angle DZM. And the angle QAG is also equal to the angle MDZ.
δύο δὴ τρίγωνά ἐστι τὰ ΜΔΖ, ΘΑΓ δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν τὴν ΘΑ τῇ ΜΔ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ.
Thus there are two triangles MDZ, QAG having two angles equal to two angles, each to each, and one side equal to one side, namely the side QA subtending one of the equal angles to MD; therefore they will also have the remaining sides equal to the remaining sides, each to each.
ἴση ἄρα ἐστὶν ἡ ΑΓ τῇ ΔΖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΑΒ τῇ ΔΕ ἐστιν ἴση
Therefore AG is equal to DZ. Similarly indeed we shall show that AB is also equal to DE.

Notes

  1. §11.prop.35#1ἐπὶ τὰς ἐξ ἀρχῆς γωνίας — Although literally meaning "toward the original angles," in this context it means joining the straight lines to the vertices of the original angles. Therefore, the translations supply "to the vertices of" or translate accordingly.
  2. §11.prop.35#1κατὰ τὰ Ν, Λ — Given that the perpendiculars are HL and MN, the perpendicular from H should meet at L, and that from M at N. However, the textual tradition has "at N, L," which reverses the order.
  3. §11.prop.35#1ΘΓ, ΓΒ, ΜΖ, ΖΕ — Based on the context and construction, the straight lines to be joined should be QG, QB, MZ, ME, but the text has GB, ZE instead. The translations follow the literal reading of the text.

Cite this passage

Euclid, Elements §11.prop.35#1. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:11.prop.35%231

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