---
title: "Elements · Στοιχεῖα"
author: "Euclid"
translator: "Thomas Little Heath"
languages: [grc, en]
shelf: "Classical Greek"
section: 40 of 49
url: https://scriptura.si/works/euclid-elements/40/
cite: "Euclid, tr. Thomas Little Heath, Elements · Στοιχεῖα, section 40. Scriptura (scriptura.si), https://scriptura.si/works/euclid-elements/40/"
---

# Elements · Στοιχεῖα

*Euclid, tr. Thomas Little Heath*

The thirteen books of geometry and number theory that taught mathematics for two thousand years, from the definition of a point to the five regular solids.

> Greek: ed. Johan Ludvig Heiberg, from the Perseus Digital Library (canonical-greekLit, tlg1799.tlg001, CC BY-SA 4.0).

> English: Thomas Little Heath, from the Perseus Digital Library (tlg1799.tlg001.perseus-eng2). Notes are not included.

> Rows follow the references the two texts share (books, chapters and sections), cut where both reach the end of a sentence.

## 40. Book XI · Propositions · 1–17

Source: https://scriptura.si/works/euclid-elements/40/ · cite a paragraph as https://scriptura.si/works/euclid-elements/40/#p-N

[¶0] 1. A part of a straight line cannot be in the plane of reference and a part in a plane more elevated. For, if possible, let a part AB of the straight line ABC be in the plane of reference, and a part BC in a plane more elevated. There will then be in the plane of reference some straight line continuous with AB in a straight line. Let it be BD; therefore AB is a common segment of the two straight lines ABC, ABD: which is impossible, inasmuch as, if we describe a circle with centre B and distance AB, the diameters will cut off unequal circumferences of the circle. Therefore a part of a straight line cannot be in the plane of reference, and a part in a plane more elevated. Q. E. D.

> Greek: 1. εὐθείας γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑποκειμένῳ, ἐπιπέδῳ, μέρος δέ τι ἐν μετεωροτέρῳ. εἰ γὰρ δυνατόν, εὐθείας γραμμῆς τῆς ΑΒΓ μέρος μέν τι τὸ ΑΒ ἔστω ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, μέρος δέ τι τὸ ΒΓ ἐν μετεωροτέρῳ. ἔσται δή τις τῇ ΑΒ συνεχὴς εὐθεῖα ἐπʼ εὐθείας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ. ἔστω ἡ ΒΔ· δύο ἄρα εὐθειῶν τῶν ΑΒΓ, ΑΒΔ κοινὸν τμῆμά ἐστιν ἡ ΑΒ· ὅπερ ἐστὶν ἀδύνατον, ἐπειδήπερ ἐὰν κέντρῳ τῷ Β καὶ διαστήματι τῷ ΑΒ κύκλον γράψωμεν, αἱ διάμετροι ἀνίσους ἀπολήψονται τοῦ κύκλου περιφερείας. εὐθείας ἄρα γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν μετεωροτέρῳ· ὅπερ ἔδει δεῖξαι.

[¶1] 2. If two straight lines cut one another, they are in one plane, and every triangle is in one plane. For let the two straight lines AB, CD cut one another at the point E; I say that AB, CD are in one plane, and every triangle is in one plane. For let points F, G be taken at random on EC, EB, let CB, FG be joined, and let FH, GK be drawn across; I say first that the triangle ECB is in one plane. For, if part of the triangle ECB, either FHC or GBK, is in the plane of reference, and the rest in another, a part also of one of the straight lines EC, EB will be in the plane of reference, and a part in another. But, if the part FCBG of the triangle ECB be in the plane of reference, and the rest in another, a part also of both the straight lines EC, EB will be in the plane of reference and a part in another: which was proved absurd. [] Therefore the triangle ECB is in one plane. But, in whatever plane the triangle ECB is, in that plane also is each of the straight lines EC, EB, and, in whatever plane each of the straight lines EC, EB is, in that plane are AB, CD also. [] Therefore the straight lines AB, CD are in one plane, and every triangle is in one plane. Q. E. D.

> Greek: 2. ἐὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, ἐν ἑνί εἰσιν ἐπιπέδῳ, καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον· λέγω, ὅτι αἱ ΑΒ, ΓΔ ἐν ἑνί εἰσιν ἐπιπέδῳ, καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. εἰλήφθω γὰρ ἐπὶ τῶν ΕΓ, ΕΒ τυχόντα σημεῖα τὰ Ζ, Η, καὶ ἐπεζεύχθωσαν αἱ ΓΒ, ΖΗ, καὶ διήχθωσαν αἱ ΖΘ, ΗΚ· λέγω πρῶτον, ὅτι τὸ ΕΓΒ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. εἰ γάρ ἐστι τοῦ ΕΓΒ τριγώνου μέρος ἤτοι τὸ ΖΘΓ ἢ τὸ ΗΒΚ ἐν τῷ ὑποκειμένῳ [ἐπιπέδῳ], τὸ δὲ λοιπὸν ἐν ἄλλῳ, ἔσται καὶ μιᾶς τῶν ΕΓ, ΕΒ εὐθειῶν μέρος μέν τι ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν ἄλλῳ. εἰ δὲ τοῦ ΕΓΒ τριγώνου τὸ ΖΓΒΗ μέρος ᾖ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ λοιπὸν ἐν ἄλλῳ, ἔσται καὶ ἀμφοτέρων τῶν ΕΓ, ΕΒ εὐθειῶν μέρος μέν τι ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν ἄλλῳ· ὅπερ ἄτοπον ἐδείχθη. τὸ ἄρα ΕΓΒ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ. ἐν ᾧ δέ ἐστι τὸ ΕΓΒ τρίγωνον, ἐν τούτῳ καὶ ἑκατέρα τῶν ΕΓ, ΕΒ, ἐν ᾧ δὲ ἑκατέρα τῶν ΕΓ, ΕΒ, ἐν τούτῳ καὶ αἱ ΑΒ, ΓΔ. αἱ ΑΒ, ΓΔ ἄρα εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ, καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ· ὅπερ ἔδει δεῖξαι.

[¶2] 3. If two planes cut one another, their common section is a straight line. For let the two planes AB, BC cut one another, and let the line DB be their common section; I say that the line DB is a straight line. For, if not, from D to B let the straight line DEB be joined in the plane AB, and in the plane BC the straight line DFB. Then the two straight lines DEB, DFB will have the same extremities, and will clearly enclose an area: which is absurd. Therefore DEB, DFB are not straight lines. Similarly we can prove that neither will there be any other straight line joined from D to B except DB the common section of the planes AB, BC. Therefore etc. Q. E. D.

> Greek: 3. ἐὰν δύο ἐπίπεδα τέμνῃ ἄλληλα, ἡ κοινὴ αὐτῶν τομὴ εὐθεῖά ἐστιν. δύο γὰρ ἐπίπεδα τὰ ΑΒ, ΒΓ τεμνέτω ἄλληλα, κοινὴ δὲ αὐτῶν τομὴ ἔστω ἡ ΔΒ γραμμή· λέγω, ὅτι ἡ ΔΒ γραμμὴ εὐθεῖά ἐστιν. εἰ γὰρ μή, ἐπεζεύχθω ἀπὸ τοῦ Δ ἐπὶ τὸ Β ἐν μὲν τῷ ΑΒ ἐπιπέδῳ εὐθεῖα ἡ ΔΕΒ, ἐν δὲ τῷ ΒΓ ἐπιπέδῳ εὐθεῖα ἡ ΔΖΒ. ἔσται δὴ δύο εὐθειῶν τῶν ΔΕΒ, ΔΖΒ τὰ αὐτὰ πέρατα, καὶ περιέξουσι δηλαδὴ χωρίον· ὅπερ ἄτοπον. οὐκ ἄρα αἱ ΔΕΒ, ΔΖΒ εὐθεῖαί εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἄλλη τις ἀπὸ τοῦ Δ ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἔσται πλὴν τῆς ΔΒ κοινῆς τομῆς τῶν ΑΒ, ΒΓ ἐπιπέδων. ἐὰν ἄρα δύο ἐπίπεδα τέμνῃ ἄλληλα, ἡ κοινὴ αὐτῶν τομὴ εὐθεῖά ἐστιν· ὅπερ ἔδει δεῖξαι.

[¶3] 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. 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 lton EFgt. Now, since the two straight lines AE, ED are equal to the two straight lines CE, EB, and contain equal angles, [] therefore the base AD is equal to the base CB, and the triangle AED will be equal to the triangle CEB; [] so that the angle DAE is also equal to the angle EBC. But the angle AEG is also equal to the angle BEH; [] 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. [] 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. [] 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. [] 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. [] 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. [] 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; [] 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. Therefore etc. Q. E. D.

> Greek: 4. ἐὰν εὐθεῖα δύο εὐθείαις τεμνούσαις ἀλλήλας πρὸς ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, καὶ τῷ διʼ 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 διʼ αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ὅπερ ἔδει δεῖξαι.

[¶4] 5. If a straight line be set up at right angles to three straight lines which meet one another, at their common point of section, the three straight lines are in one plane. For let a straight line AB be set up at right angles to the three straight lines BC, BD, BE, at their point of meeting at B; I say that BC, BD, BE are in one plane. For suppose they are not, but, if possible, let BD, BE be in the plane of reference and BC in one more elevated; let the plane through AB, BC be produced; it will thus make, as common section in the plane of reference, a straight line. [] Let it make BF. Therefore the three straight lines AB, BC, BF are in one plane, namely that drawn through AB, BC. Now, since AB is at right angles to each of the straight lines BD, BE, therefore AB is also at right angles to the plane through BD, BE. [] But the plane through BD, BE is the plane of reference; therefore AB is at right angles to the plane of reference. Thus AB will also make right angles with all the straight lines which meet it and are in the plane of reference. [] But BF which is in the plane of reference meets it; therefore the angle ABF is right. But, by hypothesis, the angle ABC is also right; therefore the angle ABF is equal to the angle ABC. And they are in one plane: which is impossible. Therefore the straight line BC is not in a more elevated plane; therefore the three straight lines BC, BD, BE are in one plane. Therefore, if a straight line be set up at right angles to three straight lines, at their point of meeting, the three straight lines are in one plane. Q. E. D.

> Greek: 5. ἐὰν εὐθεῖα τρισὶν εὐθείαις ἁπτομέναις ἀλλήλων πρὸς ὀρθὰς ἐπὶ τῆς κοινῆς τομῆς ἐπισταθῇ, αἱ τρεῖς εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ. 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. 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. 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. 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. ὑπόκειται δὲ καὶ ἡ ὑπὸ ΑΒΓ ὀρθή· ἴση ἄρα ἡ ὑπὸ ΑΒΖ γωνία τῇ ὑπὸ ΑΒΓ. καί εἰσιν ἐν ἑνὶ ἐπιπέδῳ· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ΒΓ εὐθεῖα ἐν μετεωροτέρῳ ἐστὶν ἐπιπέδῳ· αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΒΓ, ΒΔ, ΒΕ ἐν ἑνί εἰσιν ἐπιπέδῳ. ἐὰν ἄρα εὐθεῖα τρισὶν εὐθείαις ἁπτομέναις ἀλλήλων ἐπὶ τῆς ἁφῆς πρὸς ὀρθὰς ἐπισταθῇ, αἱ τρεῖς εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ· ὅπερ ἔδει δεῖξαι.

[¶5] 6. If two straight lines be at right angles to the same plane, the straight lines will be parallel. For let the two straight lines AB, CD be at right angles to the plane of reference; I say that AB is parallel to CD. For let them meet the plane of reference at the points B, D, let the straight line BD be joined, let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined. Now, since AB is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [] But each of the straight lines BD, BE is in the plane of reference and meets AB; therefore each of the angles ABD, ABE is right. For the same reason each of the angles CDB, CDE is also right. And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and they include right angles; therefore the base AD is equal to the base BE. [] And, since AB is equal to DE, while AD is also equal to BE, the two sides AB, BE are equal to the two sides ED, DA; and AE is their common base; therefore the angle ABE is equal to the angle EDA. [] But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to DA. But it is also at right angles to each of the straight lines BD, DC; therefore ED is set up at right angles to the three straight lines BD, DA, DC at their point of meeting; therefore the three straight lines BD, DA, DC are in one plane. [] But, in whatever plane DB, DA are, in that plane is AB also, for every triangle is in one plane; [] therefore the straight lines AB, BD, DC are in one plane. And each of the angles ABD, BDC is right; therefore AB is parallel to CD. [] Therefore etc. Q. E. D.

> Greek: 6. ἐὰν δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, παράλληλοι ἔσονται αἱ εὐθεῖαι. 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 [ἄρα] τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. ἅπτεται δὲ τῆς ΑΒ ἑκατέρα τῶν ΒΔ, ΒΕ οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ· ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΑΒΔ, ΑΒΕ γωνιῶν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΔΒ, ΓΔΕ ὀρθή ἐστιν. 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. 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. ὀρθὴ δὲ ἡ ὑπὸ ΑΒΕ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΕΔΑ· ἡ ΕΔ ἄρα πρὸς τὴν ΔΑ ὀρθή ἐστιν. ἔστι δὲ καὶ πρὸς ἑκατέραν τῶν ΒΔ, ΔΓ ὀρθή. 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. 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. 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.

[¶6] 7. If two straight lines be parallel and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallel straight lines. Let AB, CD be two parallel straight lines, and let points E, F be taken at random on them respectively; I say that the straight line joining the points E, F is in the same plane with the parallel straight lines. For suppose it is not, but, if possible, let it be in a more elevated plane as EGF, and let a plane be drawn through EGF; it will then make, as section in the plane of reference, a straight line. [] Let it make it, as EF; therefore the two straight lines EGF, EF will enclose an area: which is impossible. Therefore the straight line joined from E to F is not in a plane more elevated; therefore the straight line joined from E to F is in the plane through the parallel straight lines AB, CD. Therefore etc. Q. E. D.

> Greek: 7. ἐὰν ὦσι δύο εὐθεῖαι παράλληλοι, ληφθῇ δὲ ἐφʼ ἑκατέρας αὐτῶν τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις. ἔστωσαν δύο εὐθεῖαι παράλληλοι αἱ ΑΒ, ΓΔ, καὶ εἰλήφθω ἐφʼ ἑκατέρας αὐτῶν τυχόντα σημεῖα τὰ Ε, Ζ· λέγω, ὅτι ἡ ἐπὶ τὰ Ε, Ζ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις. μὴ γάρ, ἀλλʼ εἰ δυνατόν, ἔστω ἐν μετεωροτέρῳ ὡς ἡ ΕΗΖ, καὶ διήχθω διὰ τῆς ΕΗΖ ἐπίπεδον· τομὴν δὴ ποιήσει ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν. ποιείτω ὡς τὴν ΕΖ· δύο ἄρα εὐθεῖαι αἱ ΕΗΖ, ΕΖ χωρίον περιέξουσιν· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Ε ἐπὶ τὸ Ζ ἐπιζευγνυμένη εὐθεῖα ἐν μετεωροτέρῳ ἐστὶν ἐπιπέδῳ· ἐν τῷ διὰ τῶν ΑΒ, ΓΔ ἄρα παραλλήλων ἐστὶν ἐπιπέδῳ ἡ ἀπὸ τοῦ Ε ἐπὶ τὸ Ζ ἐπιζευγνυμένη εὐθεῖα. ἐὰν ἄρα ὦσι δύο εὐθεῖαι παράλληλοι, ληφθῇ δὲ ἐφʼ ἑκατέρας αὐτῶν τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις· ὅπερ ἔδει δεῖξαι.

[¶7] 8. If two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane. Let AB, CD be two parallel straight lines, and let one of them, AB, be at right angles to the plane of reference; I say that the remaining one, CD, will also be at right angles to the same plane. For let AB, CD meet the plane of reference at the points B, D, and let BD be joined; therefore AB, CD, BD are in one plane. [] Let DE be drawn, in the plane of reference, at right angles to BD, let DE be made equal to AB, and let BE, AE, AD be joined. Now, since AB is at right angles to the plane of reference, therefore AB is also at right angles to all the straight lines which meet it and are in the plane of reference; [] therefore each of the angles ABD, ABE is right. And, since the straight line BD has fallen on the parallels AB, CD, therefore the angles ABD, CDB are equal to two right angles. [] But the angle ABD is right; therefore the angle CDB is also right; therefore CD is at right angles to BD. And, since AB is equal to DE, and BD is common, the two sides AB, BD are equal to the two sides ED, DB; and the angle ABD is equal to the angle EDB, for each is right; therefore the base AD is equal to the base BE. And, since AB is equal to DE, and BE to AD, the two sides AB, BE are equal to the two sides ED, DA respectively, and AE is their common base; therefore the angle ABE is equal to the angle EDA. But the angle ABE is right; therefore the angle EDA is also right; therefore ED is at right angles to AD. But it is also at right angles to DB; therefore ED is also at right angles to the plane through BD, DA. [] Therefore ED will also make right angles with all the straight lines which meet it and are in the plane through BD, DA. But DC is in the plane through BD, DA, inasmuch as AB, BD are in the plane through BD, DA, [] and DC is also in the plane in which AB, BD are. Therefore ED is at right angles to DC, so that CD is also at right angles to DE. But CD is also at right angles to BD. Therefore CD is set up at right angles to the two straight lines DE, DB which cut one another, from the point of section at D; so that CD is also at right angles to the plane through DE, DB. [] But the plane through DE, DB is the plane of reference; therefore CD is at right angles to the plane of reference. Therefore etc. Q. E. D.

> Greek: 8. ἐὰν ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ ἑτέρα αὐτῶν ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται. ἔστωσαν δύο εὐθεῖαι παράλληλοι αἱ ΑΒ, ΓΔ, ἡ δὲ ἑτέρα αὐτῶν ἡ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστω· λέγω, ὅτι καὶ ἡ λοιπὴ ἡ ΓΔ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται. συμβαλλέτωσαν γὰρ αἱ ΑΒ, ΓΔ τῷ ὑποκειμένῳ ἐπιπέδῳ κατὰ τὰ Β, Δ σημεῖα, καὶ ἐπεζεύχθω ἡ ΒΔ· αἱ ΑΒ, ΓΔ, ΒΔ ἄρα ἐν ἑνί εἰσιν ἐπιπέδῳ. 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 ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν ἡ ΑΒ· ὀρθὴ ἄρα [ἐστὶν] ἑκατέρα τῶν ὑπὸ ΑΒΔ, ΑΒΕ γωνιῶν. καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπέπτωκεν ἡ ΒΔ, αἱ ἄρα ὑπὸ ΑΒΔ, ΓΔΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΑΒΔ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΓΔΒ· ἡ ΓΔ ἄρα πρὸς τὴν ΒΔ ὀρθή ἐστιν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΔΕ, κοινὴ δὲ ἡ ΒΔ, δύο δὴ αἱ ΑΒ, ΒΔ δυσὶ ταῖς ΕΔ, ΔΒ ἴσαι εἰσίν· καὶ γωνία ἡ ὑπὸ ΑΒΔ γωνίᾳ τῇ ὑπὸ ΕΔΒ ἴση· ὀρθὴ γὰρ ἑκατέρα· βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΕ ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΒΕ τῇ ΑΔ, δύο δὴ αἱ ΑΒ, ΒΕ δυσὶ ταῖς ΕΔ, ΔΑ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ. καὶ βάσις αὐτῶν κοινὴ ἡ ΑΕ· γωνία ἄρα ἡ ὑπὸ ΑΒΕ γωνίᾳ τῇ ὑπὸ ΕΔΑ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΑΒΕ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΕΔΑ· ἡ ΕΔ ἄρα πρὸς τὴν ΑΔ ὀρθή ἐστιν. ἔστι δὲ καὶ πρὸς τὴν ΔΒ ὀρθή· ἡ ΕΔ ἄρα καὶ τῷ διὰ τῶν ΒΔ, ΔΑ ἐπιπέδῳ ὀρθή ἐστιν. καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας ἡ ΕΔ. ἐν δὲ τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ ἐστὶν ἡ ΔΓ, ἐπειδήπερ ἐν τῷ διὰ τῶν ΒΔΑ ἐπιπέδῳ εἰσὶν αἱ ΑΒ, ΒΔ, ἐν ᾧ δὲ αἱ ΑΒ, ΒΔ, ἐν τούτῳ ἐστὶ καὶ ἡ ΔΓ. ἡ ΕΔ ἄρα τῇ ΔΓ πρὸς ὀρθάς ἐστιν· ὥστε καὶ ἡ ΓΔ τῇ ΔΕ πρὸς ὀρθάς ἐστιν. ἔστι δὲ καὶ ἡ ΓΔ τῇ ΒΔ πρὸς ὀρθάς. ἡ ΓΔ ἄρα δύο εὐθείαις τεμνούσαις ἀλλήλας ταῖς ΔΕ, ΔΒ ἀπὸ τῆς κατὰ τὸ Δ τομῆς πρὸς ὀρθὰς ἐφέστηκεν· ὥστε ἡ ΓΔ καὶ τῷ διὰ τῶν ΔΕ, ΔΒ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ τῶν ΔΕ, ΔΒ ἐπίπεδον τὸ ὑποκείμενόν ἐστιν· ἡ ΓΔ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. ἐὰν ἄρα ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ μία αὐτῶν ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ὅπερ ἔδει δεῖξαι.

[¶8] 9. Straight lines which are parallel to the same straight line and are not in the same plane with it are also parallel to one another. For let each of the straight lines AB, CD be parallel to EF, not being in the same plane with it; I say that AB is parallel to CD. For let a point G be taken at random on EF, and from it let there be drawn GH, in the plane through EF, AB, at right angles to EF, and GK in the plane through FE, CD again at right angles to EF. Now, since EF is at right angles to each of the straight lines GH, GK, therefore EF is also at right angles to the plane through GH, GK. [] And EF is parallel to AB; therefore AB is also at right angles to the plane through HG, GK. [] For the same reason CD is also at right angles to the plane through HG, GK; therefore each of the straight lines AB, CD is at right angles to the plane through HG, GK. But if two straight lines be at right angles to the same plane, the straight lines are parallel; [] therefore AB is parallel to CD.

> Greek: 9. αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ μὴ οὖσαι αὐτῇ ἐν τῷ αὐτῷ ἐπιπέδῳ καὶ ἀλλήλαις εἰσὶ παράλληλοι. ἔστω γὰρ ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος μὴ οὖσαι αὐτῇ ἐν τῷ αὐτῷ ἐπιπέδῳ· λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. εἰλήφθω γὰρ ἐπὶ τῆς ΕΖ τυχὸν σημεῖον τὸ Η, καὶ ἀπʼ αὐτοῦ τῇ ΕΖ ἐν μὲν τῷ διὰ τῶν ΕΖ, ΑΒ ἐπιπέδῳ πρὸς ὀρθὰς ἤχθω ἡ ΗΘ, ἐν δὲ τῷ διὰ τῶν ΖΕ, ΓΔ τῇ ΕΖ πάλιν πρὸς ὀρθὰς ἤχθω ἡ ΗΚ. καὶ ἐπεὶ ἡ ΕΖ πρὸς ἑκατέραν τῶν ΗΘ, ΗΚ ὀρθή ἐστιν, ἡ ΕΖ ἄρα καὶ τῷ διὰ τῶν ΗΘ, ΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. καί ἐστιν ἡ ΕΖ τῇ ΑΒ παράλληλος· καὶ ἡ ΑΒ ἄρα τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΔ τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν· ἑκατέρα ἄρα τῶν ΑΒ, ΓΔ τῷ διὰ τῶν ΘΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. ἐὰν δὲ δύο εὐθεῖαι τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν, παράλληλοί εἰσιν αἱ εὐθεῖαι· παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ· ὅπερ ἔδει δεῖξαι.

[¶9] 10. If two straight lines meeting one another be parallel to two straight lines meeting one another not in the same plane, they will contain equal angles. For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not in the same plane; I say that the angle ABC is equal to the angle DEF. For let BA, BC, ED, EF be cut off equal to one another, and let AD, CF, BE, AC, DF be joined. Now, since BA is equal and parallel to ED, therefore AD is also equal and parallel to BE. [] For the same reason CF is also equal and parallel to BE. Therefore each of the straight lines AD, CF is equal and parallel to BE. But straight lines which are parallel to the same straight line and are not in the same plane with it are parallel to one another; [] therefore AD is parallel and equal to CF. And AC, DF join them; therefore AC is also equal and parallel to DF. [] Now, since the two sides AB, BC are equal to the two sides DE, EF, and the base AC is equal to the base DF, therefore the angle ABC is equal to the angle DEF. [] Therefore etc.

> Greek: 10. ἐὰν δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἴσας γωνίας περιέξουσιν. δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΒΓ ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας τὰς ΔΕ, ΕΖ ἁπτομένας ἀλλήλων ἔστωσαν μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ· λέγω, ὅτι ἴση ἐστὶν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ. Ἀπειλήφθωσαν γὰρ αἱ ΒΑ, ΒΓ, ΕΔ, ΕΖ ἴσαι ἀλλήλαις, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΓΖ, ΒΕ, ΑΓ, ΔΖ. καὶ ἐπεὶ ἡ ΒΑ τῇ ΕΔ ἴση ἐστὶ καὶ παράλληλος, καὶ ἡ ΑΔ ἄρα τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΖ τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος· ἑκατέρα ἄρα τῶν ΑΔ, ΓΖ τῇ ΒΕ ἴση ἐστὶ καὶ παράλληλος. αἱ δὲ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ μὴ οὖσαι αὐτῇ ἐν τῷ αὐτῷ ἐπιπέδῳ καὶ ἀλλήλαις εἰσὶ παράλληλοι· παράλληλος ἄρα ἐστὶν ἡ ΑΔ τῇ ΓΖ καὶ ἴση. καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΑΓ, ΔΖ· καὶ ἡ ΑΓ ἄρα τῇ ΔΖ ἴση ἐστὶ καὶ παράλληλος. καὶ ἐπεὶ δύο αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσίν, καὶ βάσις ἡ ΑΓ βάσει τῇ ΔΖ ἴση, γωνία ἄρα ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. ἐὰν ἄρα δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἴσας γωνίας περιέξουσιν· ὅπερ ἔδει δεῖξαι.

[¶10] 11. From a given elevated point to draw a straight line perpendicular to a given plane. Let A be the given elevated point, and the plane of reference the given plane; thus it is required to draw from the point A a straight line perpendicular to the plane of reference. Let any straight line BC be drawn, at random, in the plane of reference, and let AD be drawn from the point A perpendicular to BC. [] If then AD is also perpendicular to the plane of reference, that which was enjoined will have been done. But, if not, let DE be drawn from the point D at right angles to BC and in the plane of reference, [] let AF be drawn from A perpendicular to DE, [] and let GH be drawn through the point F parallel to BC. [] Now, since BC is at right angles to each of the straight lines DA, DE, therefore BC is also at right angles to the plane through ED, DA. [] And GH is parallel to it; but, if two straight lines be parallel, and one of them be at right angles to any plane, the remaining one will also be at right angles to the same plane; [] therefore GH is also at right angles to the plane through ED, DA. Therefore GH is also at right angles to all the straight lines which meet it and are in the plane through ED, DA. [] But AF meets it and is in the plane through ED, DA; therefore GH is at right angles to FA, so that FA is also at right angles to GH. But AF is also at right angles to DE; therefore AF is at right angles to each of the straight lines GH, DE. But, if a straight line be set up at right angles to two straight lines which cut one another, at the point of section, it will also be at right angles to the plane through them; [] therefore FA is at right angles to the plane through ED, GH. But the plane through ED, GH is the plane of reference; therefore AF is at right angles to the plane of reference. Therefore from the given elevated point A the straight line AF has been drawn perpendicular to the plane of reference. Q. E. F.

> Greek: 11. ἀπὸ τοῦ δοθέντος σημείου μετεώρου ἐπὶ τὸ δοθὲν ἐπίπεδον κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. ἔστω τὸ μὲν δοθὲν σημεῖον μετέωρον τὸ Α, τὸ δὲ δοθὲν ἐπίπεδον τὸ ὑποκείμενον· δεῖ δὴ ἀπὸ τοῦ Α σημείου ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. διήχθω γάρ τις ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖα, ὡς ἔτυχεν, ἡ ΒΓ, καὶ ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ. εἰ μὲν οὖν ἡ ΑΔ κάθετός ἐστι καὶ ἐπὶ τὸ ὑποκείμενον ἐπίπεδον, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ οὔ, ἤχθω ἀπὸ τοῦ Δ σημείου τῇ ΒΓ ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΔΕ, καὶ ἤχθω ἀπὸ τοῦ Α ἐπὶ τὴν ΔΕ κάθετος ἡ ΑΖ, καὶ διὰ τοῦ Ζ σημείου τῇ ΒΓ παράλληλος ἤχθω ἡ ΗΘ. καὶ ἐπεὶ ἡ ΒΓ ἑκατέρᾳ τῶν ΔΑ, ΔΕ πρὸς ὀρθάς ἐστιν, ἡ ΒΓ ἄρα καὶ τῷ διὰ τῶν ΕΔΑ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. καί ἐστιν αὐτῇ παράλληλος ἡ ΗΘ· ἐὰν δὲ ὦσι δύο εὐθεῖαι παράλληλοι, ἡ δὲ μία αὐτῶν ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ λοιπὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· καὶ ἡ ΗΘ ἄρα τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ ὀρθή ἐστιν ἡ ΗΘ. ἅπτεται δὲ αὐτῆς ἡ ΑΖ οὖσα ἐν τῷ διὰ τῶν ΕΔ, ΔΑ ἐπιπέδῳ· ἡ ΗΘ ἄρα ὀρθή ἐστι πρὸς τὴν ΖΑ· ὥστε καὶ ἡ ΖΑ ὀρθή ἐστι πρὸς τὴν ΘΗ. ἔστι δὲ ἡ ΑΖ καὶ πρὸς τὴν ΔΕ ὀρθή· ἡ ΑΖ ἄρα πρὸς ἑκατέραν τῶν ΗΘ, ΔΕ ὀρθή ἐστιν. ἐὰν δὲ εὐθεῖα δυσὶν εὐθείαις τεμνούσαις ἀλλήλας ἐπὶ τῆς τομῆς πρὸς ὀρθὰς ἐπισταθῇ, καὶ τῷ διʼ αὐτῶν ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ἡ ΖΑ ἄρα τῷ διὰ τῶν ΕΔ, ΗΘ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ τῶν ΕΔ, ΗΘ ἐπίπεδόν ἐστι τὸ ὑποκείμενον· ἡ ΑΖ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. ἀπὸ τοῦ ἄρα δοθέντος σημείου μετεώρου τοῦ Α ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετος εὐθεῖα γραμμὴ ἦκται ἡ ΑΖ· ὅπερ ἔδει ποιῆσαι.

[¶11] 12. To set up a straight line at right angles to a given plane from a given point in it. Let the plane of reference be the given plane, and A the point in it; thus it is required to set up from the point A a straight line at right angles to the plane of reference. Let any elevated point B be conceived, from B let BC be drawn perpendicular to the plane of reference, [] and through the point A let AD be drawn parallel to BC. [] Then, since AD, CB are two parallel straight lines, while one of them, BC, is at right angles to the plane of reference, therefore the remaining one, AD, is also at right angles to the plane of reference. [] Therefore AD has been set up at right angles to the given plane from the point A in it.

> Greek: 12. τῷ δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ δοθέντος σημείου πρὸς ὀρθὰς εὐθεῖαν γραμμὴν ἀναστῆσαι. ἔστω τὸ μὲν δοθὲν ἐπίπεδον τὸ ὑποκείμενον, τὸ δὲ πρὸς αὐτῷ σημεῖον τὸ Α· δεῖ δὴ ἀπὸ τοῦ Α σημείου τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς εὐθεῖαν γραμμὴν ἀναστῆσαι. νενοήσθω τι σημεῖον μετέωρον τὸ Β, καὶ ἀπὸ τοῦ Β ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετος ἤχθω ἡ ΒΓ, καὶ διὰ τοῦ Α σημείου τῇ ΒΓ παράλληλος ἤχθω ἡ ΑΔ. ἐπεὶ οὖν δύο εὐθεῖαι παράλληλοί εἰσιν αἱ ΑΔ, ΓΒ, ἡ δὲ μία αὐτῶν ἡ ΒΓ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, καὶ ἡ λοιπὴ ἄρα ἡ ΑΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τῷ ἄρα δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ σημείου τοῦ Α πρὸς ὀρθὰς ἀνέσταται ἡ ΑΔ· ὅπερ ἔδει ποιῆσαι.

[¶12] 13. From the same point two straight lines cannot be set up at right angles to the same plane on the same side. For, if possible, from the same point A let the two straight lines AB, AC be set up at right angles to the plane of reference and on the same side, and let a plane be drawn through BA, AC; it will then make, as section through A in the plane of reference, a straight line. [] Let it make DAE; therefore the straight lines AB, AC, DAE are in one plane. And, since CA is at right angles to the plane of reference, it will also make right angles with all the straight lines which meet it and are in the plane of reference. [] But DAE meets it and is in the plane of reference; therefore the angle CAE is right. For the same reason the angle BAE is also right; therefore the angle CAE is equal to the angle BAE. And they are in one plane: which is impossible. Therefore etc. Q. E. D.

> Greek: 13. ἀπὸ τοῦ αὐτοῦ σημείου τῷ αὐτῷ ἐπιπέδῳ δύο εὐθεῖαι πρὸς ὀρθὰς οὐκ ἀναστήσονται ἐπὶ τὰ αὐτὰ μέρη. εἰ γὰρ δυνατόν, ἀπὸ τοῦ αὐτοῦ σημείου τοῦ Α τῷ ὑποκειμένῳ ἐπιπέδῳ δύο εὐθεῖαι αἱ ΑΒ, ΑΓ πρὸς ὀρθὰς ἀνεστάτωσαν ἐπὶ τὰ αὐτὰ μέρη, καὶ διήχθω τὸ διὰ τῶν ΒΑ, ΑΓ ἐπίπεδον· τομὴν δὴ ποιήσει διὰ τοῦ Α ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ εὐθεῖαν. ποιείτω τὴν ΔΑΕ· αἱ ἄρα ΑΒ, ΑΓ, ΔΑΕ εὐθεῖαι ἐν ἑνί εἰσιν ἐπιπέδῳ. καὶ ἐπεὶ ἡ ΓΑ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. ἅπτεται δὲ αὐτῆς ἡ ΔΑΕ οὖσα ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ· ἡ ἄρα ὑπὸ ΓΑΕ γωνία ὀρθή ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΒΑΕ ὀρθή ἐστιν· ἴση ἄρα ἡ ὑπὸ ΓΑΕ τῇ ὑπὸ ΒΑΕ. καί εἰσιν ἐν ἑνὶ ἐπιπέδῳ· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἀπὸ τοῦ αὐτοῦ σημείου τῷ αὐτῷ ἐπιπέδῳ δύο εὐθεῖαι πρὸς ὀρθὰς ἀνασταθήσονται ἐπὶ τὰ αὐτὰ μέρη· ὅπερ ἔδει δεῖξαι.

[¶13] 14. Planes to which the same straight line is at right angles will be parallel. For let any straight line AB be at right angles to each of the planes CD, EF; I say that the planes are parallel. For, if not, they will meet when produced. Let them meet; they will then make, as common section, a straight line. [] Let them make GH; let a point K be taken at random on GH, and let AK, BK be joined. Now, since AB is at right angles to the plane EF, therefore AB is also at right angles to BK which is a straight line in the plane EF produced; [] therefore the angle ABK is right. For the same reason the angle BAK is also right. Thus, in the triangle ABK, the two angles ABK, BAK are equal to two right angles: which is impossible. [] Therefore the planes CD, EF will not meet when produced; therefore the planes CD, EF are parallel. [] Therefore planes to which the same straight line is at right angles are parallel. Q. E. D.

> Greek: 14. πρὸς ἃ ἐπίπεδα ἡ αὐτὴ εὐθεῖα ὀρθή ἐστιν, παράλληλα ἔσται τὰ ἐπίπεδα. εὐθεῖα γάρ τις ἡ ΑΒ πρὸς ἑκάτερον τῶν ΓΔ, ΕΖ ἐπιπέδων πρὸς ὀρθὰς ἔστω· λέγω, ὅτι παράλληλά ἐστι τὰ ἐπίπεδα. εἰ γὰρ μή, ἐκβαλλόμενα συμπεσοῦνται. συμπιπτέτωσαν· ποιήσουσι δὴ κοινὴν τομὴν εὐθεῖαν. ποιείτωσαν τὴν ΗΘ, καὶ εἰλήφθω ἐπὶ τῆς ΗΘ τυχὸν σημεῖον τὸ Κ, καὶ ἐπεζεύχθωσαν αἱ ΑΚ, ΒΚ. καὶ ἐπεὶ ἡ ΑΒ ὀρθή ἐστι πρὸς τὸ ΕΖ ἐπίπεδον, καὶ πρὸς τὴν ΒΚ ἄρα εὐθεῖαν οὖσαν ἐν τῷ ΕΖ ἐκβληθέντι ἐπιπέδῳ ὀρθή ἐστιν ἡ ΑΒ· ἡ ἄρα ὑπὸ ΑΒΚ γωνία ὀρθή ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΒΑΚ ὀρθή ἐστιν. τριγώνου δὴ τοῦ ΑΒΚ αἱ δύο γωνίαι αἱ ὑπὸ ΑΒΚ, ΒΑΚ δυσὶν ὀρθαῖς εἰσιν ἴσαι· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΓΔ, ΕΖ ἐπίπεδα ἐκβαλλόμενα συμπεσοῦνται· παράλληλα ἄρα ἐστὶ τὰ ΓΔ, ΕΖ ἐπίπεδα. πρὸς ἃ ἐπίπεδα ἄρα ἡ αὐτὴ εὐθεῖα ὀρθή ἐστιν, παράλληλά ἐστι τὰ ἐπίπεδα· ὅπερ ἔδει δεῖξαι.

[¶14] 15. If two straight lines meeting one another be parallel to two straight lines meeting one another, not being in the same plane, the planes through them are parallel. For let the two straight lines AB, BC meeting one another be parallel to the two straight lines DE, EF meeting one another, not being in the same plane; I say that the planes produced through AB, BC and DE, EF will not meet one another. For let BG be drawn from the point B perpendicular to the plane through DE, EF [], and let it meet the plane at the point G; through G let GH be drawn parallel to ED, and GK parallel to EF. [] Now, since BG is at right angles to the plane through DE, EF, therefore it will also make right angles with all the straight lines which meet it and are in the plane through DE, EF. [] But each of the straight lines GH, GK meets it and is in the plane through DE, EF; therefore each of the angles BGH, BGK is right. And, since BA is parallel to GH, [] therefore the angles GBA, BGH are equal to two right angles. [] But the angle BGH is right; therefore the angle GBA is also right; therefore GB is at right angles to BA. For the same reason GB is also at right angles to BC. Since then the straight line GB is set up at right angles to the two straight lines BA, BC which cut one another, therefore GB is also at right angles to the plane through BA, BC. [] But planes to which the same straight line is at right angles are parallel; [] therefore the plane through AB, BC is parallel to the plane through DE, EF. Therefore, if two straight lines meeting one another be parallel to two straight lines meeting one another, not in the same plane, the planes through them are parallel. Q. E. D.

> Greek: 15. ἐὰν δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι, παράλληλά ἐστι τὰ διʼ αὐτῶν ἐπίπεδα. δύο γὰρ εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΑΒ, ΒΓ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΔΕ, ΕΖ ἔστωσαν μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι· λέγω, ὅτι ἐκβαλλόμενα τὰ διὰ τῶν ΑΒ, ΒΓ, ΔΕ, ΕΖ ἐπίπεδα οὐ συμπεσεῖται ἀλλήλοις. ἤχθω γὰρ ἀπὸ τοῦ Β σημείου ἐπὶ τὸ διὰ τῶν ΔΕ, ΕΖ ἐπίπεδον κάθετος ἡ ΒΗ καὶ συμβαλλέτω τῷ ἐπιπέδῳ κατὰ τὸ Η σημεῖον, καὶ διὰ τοῦ Η τῇ μὲν ΕΔ παράλληλος ἤχθω ἡ ΗΘ, τῇ δὲ ΕΖ ἡ ΗΚ. καὶ ἐπεὶ ἡ ΒΗ ὀρθή ἐστι πρὸς τὸ διὰ τῶν ΔΕ, ΕΖ ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ ὀρθὰς ποιήσει γωνίας. ἅπτεται δὲ αὐτῆς ἑκατέρα τῶν ΗΘ, ΗΚ οὖσα ἐν τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ· ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΒΗΘ, ΒΗΚ γωνιῶν. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΒΑ τῇ ΗΘ, αἱ ἄρα ὑπὸ ΗΒΑ, ΒΗΘ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΒΗΘ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΗΒΑ· ἡ ΗΒ ἄρα τῇ ΒΑ πρὸς ὀρθάς ἐστιν. διὰ τὰ αὐτὰ δὴ ἡ ΗΒ καὶ τῇ ΒΓ ἐστι πρὸς ὀρθάς. ἐπεὶ οὖν εὐθεῖα ἡ ΗΒ δυσὶν εὐθείαις ταῖς ΒΑ, ΒΓ τεμνούσαις ἀλλήλας πρὸς ὀρθὰς ἐφέστηκεν, ἡ ΗΒ ἄρα καὶ τῷ διὰ τῶν ΒΑ, ΒΓ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. [διὰ τὰ αὐτὰ δὴ ἡ ΒΗ καὶ τῷ διὰ τῶν ΗΘ, ΗΚ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν. τὸ δὲ διὰ τῶν ΗΘ, ΗΚ ἐπίπεδόν ἐστι τὸ διὰ τῶν ΔΕ, ΕΖ· ἡ ΒΗ ἄρα τῷ διὰ τῶν ΔΕ, ΕΖ ἐπιπέδῳ ἐστὶ πρὸς ὀρθάς. ἐδείχθη δὲ ἡ ΗΒ καὶ τῷ διὰ τῶν ΑΒ, ΒΓ ἐπιπέδῳ πρὸς ὀρθάς]. πρὸς ἃ δὲ ἐπίπεδα ἡ αὐτὴ εὐθεῖα ὀρθή ἐστιν, παράλληλά ἐστι τὰ ἐπίπεδα· παράλληλον ἄρα ἐστὶ τὸ διὰ τῶν ΑΒ, ΒΓ ἐπίπεδον τῷ διὰ τῶν ΔΕ, ΕΖ. ἐὰν ἄρα δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, παράλληλά ἐστι τὰ διʼ αὐτῶν ἐπίπεδα· ὅπερ ἔδει δεῖξαι.

[¶15] 16. If two parallel planes be cut by any plane, their common sections are parallel. For let the two parallel planes AB, CD be cut by the plane EFGH, and let EF, GH be their common sections; I say that EF is parallel to GH. For, if not, EF, GH will, when produced, meet either in the direction of F, H or of E, G. Let them be produced, as in the direction of F, H, and let them, first, meet at K. Now, since EFK is in the plane AB, therefore all the points on EFK are also in the plane AB. [] But K is one of the points on the straight line EFK; therefore K is in the plane AB. For the same reason K is also in the plane CD; therefore the planes AB, CD will meet when produced. But they do not meet, because they are, by hypothesis, parallel; therefore the straight lines EF, GH will not meet when produced in the direction of F, H. Similarly we can prove that neither will the straight lines EF, GH meet when produced in the direction of E, G. But straight lines which do not meet in either direction are parallel. [] Therefore EF is parallel to GH. Therefore etc. Q. E. D.

> Greek: 16. ἐὰν δύο ἐπίπεδα παράλληλα ὑπὸ ἐπιπέδου τινὸς τέμνηται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν. δύο γὰρ ἐπίπεδα παράλληλα τὰ ΑΒ, ΓΔ ὑπὸ ἐπιπέδου τοῦ ΕΖΗΘ τεμνέσθω, κοιναὶ δὲ αὐτῶν τομαὶ ἔστωσαν αἱ ΕΖ, ΗΘ· λέγω, ὅτι παράλληλός ἐστιν ἡ ΕΖ τῇ ΗΘ. εἰ γὰρ μή, ἐκβαλλόμεναι αἱ ΕΖ, ΗΘ ἤτοι ἐπὶ τὰ Ζ, Θ μέρη ἢ ἐπὶ τὰ Ε, Η συμπεσοῦνται. ἐκβεβλήσθωσαν ὡς ἐπὶ τὰ Ζ, Θ μέρη καὶ συμπιπτέτωσαν πρότερον κατὰ τὸ Κ. καὶ ἐπεὶ ἡ ΕΖΚ ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ, καὶ πάντα ἄρα τὰ ἐπὶ τῆς ΕΖΚ σημεῖα ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ. ἓν δὲ τῶν ἐπὶ τῆς ΕΖΚ εὐθείας σημείων ἐστὶ τὸ Κ· τὸ Κ ἄρα ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ. διὰ τὰ αὐτὰ δὴ τὸ Κ καὶ ἐν τῷ ΓΔ ἐστιν ἐπιπέδῳ· τὰ ΑΒ, ΓΔ ἄρα ἐπίπεδα ἐκβαλλόμενα συμπεσοῦνται. οὐ συμπίπτουσι δὲ διὰ τὸ παράλληλα ὑποκεῖσθαι· οὐκ ἄρα αἱ ΕΖ, ΗΘ εὐθεῖαι ἐκβαλλόμεναι ἐπὶ τὰ Ζ, Θ μέρη συμπεσοῦνται. ὁμοίως δὴ δείξομεν, ὅτι αἱ ΕΖ, ΗΘ εὐθεῖαι οὐδὲ ἐπὶ τὰ Ε, Η μέρη ἐκβαλλόμεναι συμπεσοῦνται. αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη συμπίπτουσαι παράλληλοί εἰσιν. παράλληλος ἄρα ἐστὶν ἡ ΕΖ τῇ ΗΘ. ἐὰν ἄρα δύο ἐπίπεδα παράλληλα ὑπὸ ἐπιπέδου τινὸς τέμνηται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν· ὅπερ ἔδει δεῖξαι.

[¶16] 17. If two straight lines be cut by parallel planes, they will be cut in the same ratios. For let the two straight lines AB, CD be cut by the parallel planes GH, KL, MN at the points A, E, B and C, F, D; I say that, as the straight line AE is to EB, so is CF to FD. For let AC, BD, AD be joined, let AD meet the plane KL at the point O, and let EO, OF be joined. Now, since the two parallel planes KL, MN are cut by the plane EBDO, their common sections EO, BD are parallel. [] For the same reason, since the two parallel planes GH, KL are cut by the plane AOFC, their common sections AC, OF are parallel. [id.] And, since the straight line EO has been drawn parallel to BD, one of the sides of the triangle ABD, therefore, proportionally, as AE is to EB, so is AO to OD. [] Again, since the straight line OF has been drawn parallel to AC, one of the sides of the triangle ADC, proportionally, as AO is to OD, so is CF to FD. [id.] But it was also proved that, as AO is to OD, so is AE to EB; therefore also, as AE is to EB, so is CF to FD. [] Therefore etc. Q. E. D.

> Greek: 17. ἐὰν δύο εὐθεῖαι ὑπὸ παραλλήλων ἐπιπέδων τέμνωνται εἰς τοὺς αὐτοὺς λόγους τμηθήσονται. δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ ὑπὸ παραλλήλων ἐπιπέδων τῶν ΗΘ, ΚΛ, ΜΝ τεμνέσθωσαν κατὰ τὰ Α, Ε, Β, Γ, Ζ, Δ σημεῖα· λέγω, ὅτι ἐστὶν ὡς ἡ ΑΕ εὐθεῖα πρὸς τὴν ΕΒ, οὕτως ἡ ΓΖ πρὸς τὴν ΖΔ. ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΒΔ, ΑΔ, καὶ συμβαλλέτω ἡ ΑΔ τῷ ΚΛ ἐπιπέδῳ κατὰ τὸ Ξ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΕΞ, ΞΖ. καὶ ἐπεὶ δύο ἐπίπεδα παράλληλα τὰ ΚΛ, ΜΝ ὑπὸ ἐπιπέδου τοῦ ΕΒΔΞ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ αἱ ΕΞ, ΒΔ παράλληλοί εἰσιν. διὰ τὰ αὐτὰ δὴ ἐπεὶ δύο ἐπίπεδα παράλληλα τὰ ΗΘ, ΚΛ ὑπὸ ἐπιπέδου τοῦ ΑΞΖΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ αἱ ΑΓ, ΞΖ παράλληλοί εἰσιν. καὶ ἐπεὶ τριγώνου τοῦ ΑΒΔ παρὰ μίαν τῶν πλευρῶν τὴν ΒΔ εὐθεῖα ἦκται ἡ ΕΞ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΕ πρὸς ΕΒ, οὕτως ἡ ΑΞ πρὸς ΞΔ. πάλιν ἐπεὶ τριγώνου τοῦ ΑΔΓ παρὰ μίαν τῶν πλευρῶν τὴν ΑΓ εὐθεῖα ἦκται ἡ ΞΖ, ἀνάλογόν ἐστιν ὡς ἡ ΑΞ πρὸς ΞΔ, οὕτως ἡ ΓΖ πρὸς ΖΔ. ἐδείχθη δὲ καὶ ὡς ἡ ΑΞ πρὸς ΞΔ, οὕτως ἡ ΑΕ πρὸς ΕΒ· καὶ ὡς ἄρα ἡ ΑΕ πρὸς ΕΒ, οὕτως ἡ ΓΖ πρὸς ΖΔ. ἐὰν ἄρα δύο εὐθεῖαι ὑπὸ παραλλήλων ἐπιπέδων τέμνωνται, εἰς τοὺς αὐτοὺς λόγους τμηθήσονται· ὅπερ ἔδει δεῖξαι.