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Euclid · Elements §11.prop.5-11.prop.6

Coplanar Lines at a Perpendicular and Parallel Normals

Passage 243 of 316 · Greek

Summary

In proposition 5, it is proved by contradiction that if a straight line is set up at right angles to three intersecting lines at their common point of intersection, the three lines lie in the same plane. In proposition 6, it is proved using auxiliary lines and congruent triangles that two straight lines perpendicular to the same plane are parallel to each other.

§11.prop.5ἐὰν εὐθεῖα τρισὶν εὐθείαις ἁπτομέναις ἀλλήλων πρὸς ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, αἱ τρεῖς εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ.
If a straight line be set up at right angles to three straight lines meeting one another, at their common point of intersection, the three straight lines will be in one plane.
εὐθεῖα γάρ τις ἡ ΑΒ τρισὶν εὐθείαις ταῖς ΒΓ, ΒΔ, ΒΕ πρὸς ὀρθὰς ἐπὶ τῆς κατὰ τὸ Β ἁφῆς ἐφεστάτω· λέγω, ὅτι αἱ ΒΓ, ΒΔ, ΒΕ ἐν ἑνί εἰσιν ἐπιπέδῳ.
For let a certain straight line AB be set up at right angles to three straight lines BG, BD, BE at their point of contact at B; I say that BG, BD, BE are in one plane.
μὴ γάρ, ἀλλʼ εἰ δυνατόν, ἔστωσαν αἱ μὲν ΒΔ, ΒΕ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, ἡ δὲ ΒΓ ἐν μετεωροτέρῳ, καὶ ἐκβεβλήσθω τὸ διὰ τῶν ΑΒ, ΒΓ ἐπίπεδον· κοινὴν δὴ τομὴν ποιήσει ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν.
For let it not be so, but, if possible, let BD, BE be in the underlying plane, and BG in a more elevated one, and let the plane through AB, BG be produced; indeed, it will make as a common section in the underlying plane a straight line. Let it make BZ.
ποιείτω τὴν ΒΖ. ἐν ἑνὶ ἄρα εἰσὶν ἐπιπέδῳ τῷ διηγμένῳ διὰ τῶν ΑΒ, ΒΓ αἱ τρεῖς εὐθεῖαι αἱ ΑΒ, ΒΓ, ΒΖ. καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς ἑκατέραν τῶν ΒΔ, ΒΕ, καὶ τῷ διὰ τῶν ΒΔ, ΒΕ ἄρα ἐπιπέδῳ ὀρθή ἐστιν ἡ ΑΒ. τὸ δὲ διὰ τῶν ΒΔ, ΒΕ ἐπίπεδον τὸ ὑποκείμενόν ἐστιν· ἡ ΑΒ ἄρα ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον.
Therefore the three straight lines AB, BG, BZ are in one plane, namely that drawn through AB, BG. And since AB is at right angles to each of BD, BE, therefore AB is also at right angles to the plane through BD, BE. But the plane through BD, BE is the underlying plane; therefore AB is at right angles to the underlying plane.
ὥστε καὶ πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας ἡ ΑΒ. ἅπτεται δὲ αὐτῆς ἡ ΒΖ οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ· ἡ ἄρα ὑπὸ ΑΒΖ γωνία ὀρθή ἐστιν.
So that AB will also make right angles with all the straight lines meeting it and being in the underlying plane. But BZ, being in the underlying plane, meets it; therefore the angle ABZ is a right angle.
ὑπόκειται δὲ καὶ ἡ ὑπὸ ΑΒΓ ὀρθή· ἴση ἄρα ἡ ὑπὸ ΑΒΖ γωνία τῇ ὑπὸ ΑΒΓ. καί εἰσιν ἐν ἑνὶ ἐπιπέδῳ· ὅπερ ἐστὶν ἀδύνατον.
But the angle ABG is also assumed to be right; therefore the angle ABZ is equal to the angle ABG. And they are in one plane; which is impossible.
οὐκ ἄρα ἡ ΒΓ εὐθεῖα ἐν μετεωροτέρῳ ἐστὶν ἐπιπέδῳ· αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΒΓ, ΒΔ, ΒΕ ἐν ἑνί εἰσιν ἐπιπέδῳ.
Therefore the straight line BG is not in a more elevated plane; therefore the three straight lines BG, BD, BE are in one plane.
ἐὰν ἄρα εὐθεῖα τρισὶν εὐθείαις ἁπτομέναις ἀλλήλων ἐπὶ τῆς ἁφῆς πρὸς ὀρθὰς ἐπισταθῇ, αἱ τρεῖς εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ· ὅπερ ἔδει δεῖξαι.
Therefore, if a straight line be set up at right angles to three straight lines meeting one another, at their point of contact, the three straight lines will be in one plane; which was to be proved.
§11.prop.6ἐὰν δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, παράλληλοι ἔσονται αἱ εὐθεῖαι.
If two straight lines be at right angles to the same plane, the straight lines will be parallel.
δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστωσαν· λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. συμβαλλέτωσαν γὰρ τῷ ὑποκειμένῳ ἐπιπέδῳ κατὰ τὰ Β, Δ σημεῖα, καὶ ἐπεζεύχθω ἡ ΒΔ εὐθεῖα, καὶ ἤχθω τῇ ΒΔ πρὸς ὀρθὰς ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ἡ ΔΕ, καὶ κείσθω τῇ ΑΒ ἴση ἡ ΔΕ, καὶ ἐπεζεύχθωσαν αἱ ΒΕ, ΑΕ, ΑΔ. καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον, καὶ πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας.
For let two straight lines AB, GD be at right angles to the underlying plane; I say that AB is parallel to GD. For let them meet the underlying plane at the points B, D, and let the straight line BD be joined, and let DE be drawn at right angles to BD in the underlying plane, and let DE be made equal to AB, and let BE, AE, AD be joined. And since AB is at right angles to the underlying plane, it will also make right angles with all the straight lines meeting it and being in the underlying plane.
ἅπτεται δὲ τῆς ΑΒ ἑκατέρα τῶν ΒΔ, ΒΕ οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ· ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΑΒΔ, ΑΒΕ γωνιῶν.
But each of BD, BE, being in the underlying plane, meets AB; therefore each of the angles ABD, ABE is right.
διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΔΒ, ΓΔΕ ὀρθή ἐστιν.
For the same reasons indeed, each of the angles GDB, GDE is also right.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, κοινὴ δὲ ἡ ΒΔ, δύο δὴ αἱ ΑΒ, ΒΔ δυσὶ ταῖς ΕΔ, ΔΒ ἴσαι εἰσίν· καὶ γωνίας ὀρθὰς περιέχουσιν· βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΕ ἐστιν ἴση.
And since AB is equal to DE, and BD is common, indeed the two sides AB, BD are equal to the two ED, DB; and they contain right angles; therefore the base AD is equal to the base BE.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, ἀλλὰ καὶ ἡ ΑΔ τῇ ΒΕ, δύο δὴ αἱ ΑΒ, ΒΕ δυσὶ ταῖς ΕΔ, ΔΑ ἴσαι εἰσίν· καὶ βάσις αὐτῶν κοινὴ ἡ ΑΕ· γωνία ἄρα ἡ ὑπὸ ΑΒΕ γωνίᾳ τῇ ὑπὸ ΕΔΑ ἐστιν ἴση.
And since AB is equal to DE, but also AD to BE, indeed the two sides AB, BE are equal to the two ED, DA; and their base AE is common; therefore the angle ABE is equal to the angle EDA.
ὀρθὴ δὲ ἡ ὑπὸ ΑΒΕ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΕΔΑ· ἡ ΕΔ ἄρα πρὸς τὴν ΔΑ ὀρθή ἐστιν.
But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to DA.
ἔστι δὲ καὶ πρὸς ἑκατέραν τῶν ΒΔ, ΔΓ ὀρθή.
But it is also at right angles to each of BD, DG.
ἡ ΕΔ ἄρα τρισὶν εὐθείαις ταῖς ΒΔ, ΔΑ, ΔΓ πρὸς ὀρθὰς ἐπὶ τῆς ἁφῆς ἐφέστηκεν· αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΒΔ, ΔΑ, ΔΓ ἐν ἑνί εἰσιν ἐπιπέδῳ.
Therefore ED has been set up at right angles to three straight lines BD, DA, DG at their point of contact; therefore the three straight lines BD, DA, DG are in one plane.
ἐν ᾧ δὲ αἱ ΔΒ, ΔΑ, ἐν τούτῳ καὶ ἡ ΑΒ· πᾶν γὰρ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ· αἱ ἄρα ΑΒ, ΒΔ, ΔΓ εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ.
But in that plane in which DB, DA are, AB is also; for every triangle is in one plane; therefore the straight lines AB, BD, DG are in one plane.
καί ἐστιν ὀρθὴ ἑκατέρα τῶν ὑπὸ ΑΒΔ, ΒΔΓ γωνιῶν· παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. ἐὰν ἄρα δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, παράλληλοι ἔσονται αἱ εὐθεῖαι· ὅπερ ἔδει δεῖξαι.
And each of the angles ABD, BDG is right; therefore AB is parallel to GD. Therefore, if two straight lines be at right angles to the same plane, the straight lines will be parallel; which was to be proved.

Notes

  1. §11.prop.5μὴ γάρ, ἀλλʼ εἰ δυνατόν — A classical formulaic expression in Greek mathematical texts used to initiate a proof by contradiction (reductio ad absurdum). It is translated as 'For let it not be so, but, if possible, let...', introducing the assumption of the contrary to the desired conclusion in order to derive an impossibility.
  2. §11.prop.5ἐν μετεωροτέρῳ — The comparative form of the adjective μετέωρος (elevated, in the air). In solid geometry, it is a technical usage indicating that a line or plane is raised or tilted relative to the 'underlying plane' (τὸ ὑποκείμενον ἐπίπεδον) and thus does not coincide with it.
  3. §11.prop.6ἐν ᾧ δὲ αἱ ΔΒ, ΔΑ, ἐν τούτῳ καὶ ἡ ΑΒ — The noun ἐπιπέδῳ (plane), which is the antecedent of the relative pronoun ᾧ, is omitted. The structural meaning is 'In that [plane] in which DB, DA are, in this [plane] AB is also.' This establishes that the line AB lies in the plane determined by DB and DA, relying on the immediate explanation that 'every triangle is in one plane' (since the three points A, B, and D form a triangle).

Cite this passage

Euclid, Elements §11.prop.5-11.prop.6. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:11.prop.5-11.prop.6

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