Book XI, Proposition 4

If a straight line be set up at right angles to two straight lines which cut one another, at their common point of section, it will also be at right angles to the plane through them.

Ἐὰν εὐθεῖα δύο εὐθείαις τεμνούσαις ἀλλήλας πρὸς ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, καὶ τῷ δι' αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται. Εὐθεῖα γάρ τις ἡ ΕΖ δύο εὐθείαις ταῖς ΑΒ, ΓΔ τεμνούσαις ἀλλήλας κατὰ τὸ Ε σημεῖον ἀπὸ τοῦ Ε πρὸς ὀρθὰς ἐφεστάτω: λέγω, ὅτι ἡ ΕΖ καὶ τῷ διὰ τῶν ΑΒ, ΓΔ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. Ἀπειλήφθωσαν γὰρ αἱ ΑΕ, ΕΒ, ΓΕ, ΕΔ ἴσαι ἀλλήλαις, καὶ διήχθω τις διὰ τοῦ Ε, ὡς ἔτυχεν, ἡ ΗΕΘ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΓΒ, καὶ ἔτι ἀπὸ τυχόντος τοῦ Ζ ἐπεζεύχθωσαν αἱ ΖΑ, ΖΗ, ΖΔ, ΖΓ, ΖΘ, ΖΒ. καὶ ἐπεὶ δύο αἱ ΑΕ, ΕΔ δυσὶ ταῖς ΓΕ, ΕΒ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΑΔ βάσει τῇ ΓΒ ἴση ἐστίν, καὶ τὸ ΑΕΔ τρίγωνον τῷ ΓΕΒ τριγώνῳ ἴσον ἔσται: ὥστε καὶ γωνία ἡ ὑπὸ ΔΑΕ γωνίᾳ τῇ ὑπὸ ΕΒΓ ἴση [ ἐστίν ]. ἔστι δὲ καὶ ἡ ὑπὸ ΑΕΗ γωνία τῇ ὑπὸ ΒΕΘ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΑΗΕ, ΒΕΘ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΑΕ τῇ ΕΒ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν. ἴση ἄρα ἡ μὲν ΗΕ τῇ ΕΘ, ἡ δὲ ΑΗ τῇ ΒΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΖΕ, βάσις ἄρα ἡ ΖΑ βάσει τῇ ΖΒ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΖΓ τῇ ΖΔ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΓΒ, ἔστι δὲ καὶ ἡ ΖΑ τῇ ΖΒ ἴση, δύο δὴ αἱ ΖΑ, ΑΔ δυσὶ ταῖς ΖΒ, ΒΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΖΔ βάσει τῇ ΖΓ ἐδείχθη ἴση: καὶ γωνία ἄρα ἡ ὑπὸ ΖΑΔ γωνίᾳ τῇ ὑπὸ ΖΒΓ ἴση ἐστίν. καὶ ἐπεὶ πάλιν ἐδείχθη ἡ ΑΗ τῇ ΒΘ ἴση, ἀλλὰ μὴν καὶ ἡ ΖΑ τῇ ΖΒ ἴση, δύο δὴ αἱ ΖΑ, ΑΗ δυσὶ ταῖς ΖΒ, ΒΘ ἴσαι εἰσίν. καὶ γωνία ἡ ὑπὸ ΖΑΗ ἐδείχθη ἴση τῇ ὑπὸ ΖΒΘ: βάσις ἄρα ἡ ΖΗ βάσει τῇ ΖΘ ἐστιν ἴση. καὶ ἐπεὶ πάλιν ἴση ἐδείχθη ἡ ΗΕ τῇ ΕΘ, κοινὴ δὲ ἡ ΕΖ, δύο δὴ αἱ ΗΕ, ΕΖ δυσὶ ταῖς ΘΕ, ΕΖ ἴσαι εἰσίν: καὶ βάσις ἡ ΖΗ βάσει τῇ ΖΘ ἴση: γωνία ἄρα ἡ ὑπὸ ΗΕΖ γωνίᾳ τῇ ὑπὸ ΘΕΖ ἴση ἐστίν. ὀρθὴ ἄρα ἑκατέρα τῶν ὑπὸ ΗΕΖ, ΘΕΖ γωνιῶν. ἡ ΖΕ ἄρα πρὸς τὴν ΗΘ τυχόντως διὰ τοῦ Ε ἀχθεῖσαν ὀρθή ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι ἡ ΖΕ καὶ πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. εὐθεῖα δὲ πρὸς ἐπίπεδον ὀρθή ἐστιν, ὅταν πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ αὐτῷ ἐπιπέδῳ ὀρθὰς ποιῇ γωνίας: ἡ ΖΕ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ ὑποκείμενον ἐπίπεδόν ἐστι τὸ διὰ τῶν ΑΒ, ΓΔ εὐθειῶν. ἡ ΖΕ ἄρα πρὸς ὀρθάς ἐστι τῷ διὰ τῶν ΑΒ, ΓΔ ἐπιπέδῳ. Ἐὰν ἄρα εὐθεῖα δύο εὐθείαις τεμνούσαις ἀλλήλας πρὸς ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, καὶ τῷ δι' αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται: ὅπερ ἔδει δεῖξαι.

If a straight line be set up at right angles to two straight lines which cut one another, at their common point of section, it will also be at right angles to the plane through them. For let a straight line EF be set up at right angles to the two straight lines AB, CD, which cut one another at the point E, from E; I say that EF is also at right angles to the plane through AB, CD. For let AE, EB, CE, ED be cut off equal to one another, and let any straight line GEH be drawn across through E, at random; let AD, CB be joined, and further let FA, FG, FD, FC, FH, FB be joined from the point F taken at random <on EF>. Now, since the two straight lines AE, ED are equal to the two straight lines CE, EB, and contain equal angles, [I. 15] therefore the base AD is equal to the base CB, and the triangle AED will be equal to the triangle CEB; [I. 4] so that the angle DAE is also equal to the angle EBC. But the angle AEG is also equal to the angle BEH; [I. 15] therefore AGE, BEH are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say, AE to EB; therefore they will also have the remaining sides equal to the remaining sides. [I. 26] Therefore GE is equal to EH, and AG to BH. And, since AE is equal to EB, while FE is common and at right angles, therefore the base FA is equal to the base FB. [I. 4] For the same reason FC is also equal to FD. And, since AD is equal to CB, and FA is also equal to FB, the two sides FA, AD are equal to the two sides FB, BC respectively; and the base FD was proved equal to the base FC; therefore the angle FAD is also equal to the angle FBC. [I. 8] And since, again, AG was proved equal to BH, and further FA also equal to FB, the two sides FA, AG are equal to the two sides FB, BH. And the angle FAG was proved equal to the angle FBH; therefore the base FG is equal to the base FH. [I. 4] Now since, again, GE was proved equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the base FG is equal to the base FH; therefore the angle GEF is equal to the angle HEF. [I. 8] Therefore each of the angles GEF, HEF is right. Therefore FE is at right angles to GH drawn at random through E. Similarly we can prove that FE will also make right angles with all the straight lines which meet it and are in the plane of reference. But a straight line is at right angles to a plane when it makes right angles with all the straight lines which meet it and are in that same plane; [XI. Def. 3] therefore FE is at right angles to the plane of reference. But the plane of reference is the plane through the straight lines AB, CD. Therefore FE is at right angles to the plane through AB, CD.

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