§11.prop.1εὐθείας γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑποκειμένῳ, ἐπιπέδῳ, μέρος δέ τι ἐν μετεωροτέρῳ.
A part of a straight line cannot be in the underlying plane, and another part in a more elevated place.
εἰ γὰρ δυνατόν, εὐθείας γραμμῆς τῆς ΑΒΓ μέρος μέν τι τὸ ΑΒ ἔστω ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, μέρος δέ τι τὸ ΒΓ ἐν μετεωροτέρῳ.
For, if possible, let a part AB of the straight line ABG be in the underlying plane, and a part BG in a more elevated place.
ἔσται δή τις τῇ ΑΒ συνεχὴς εὐθεῖα ἐπʼ εὐθείας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ.
There will indeed be some straight line continuous with AB and in a straight line with it in the underlying plane.
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.
Let it be BD; therefore, of the two straight lines ABG, ABD, AB is a common segment; which is impossible, since if we describe a circle with centre B and distance AB, the diameters will cut off unequal circumferences of the circle.
εὐθείας ἄρα γραμμῆς μέρος μέν τι οὐκ ἔστιν ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν μετεωροτέρῳ· ὅπερ ἔδει δεῖξαι.
Therefore, a part of a straight line is not in the underlying plane, and another part in a more elevated place; which was to be proved.
§11.prop.2ἐὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, ἐν ἑνί εἰσιν ἐπιπέδῳ, καὶ πᾶν τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ.
If two straight lines cut one another, they are in one plane, and every triangle is in one plane.
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.
For let two straight lines AB, GD cut one another at the point E; I say that AB, GD are in one plane, and every triangle is in one plane.
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.
For let arbitrary points Z, H be taken on EG, EB, and let GB, ZH be joined, and let ZQ, HK be drawn through; I say first that the triangle EGB is in one plane.
εἰ γάρ ἐστι τοῦ ΕΓΒ τριγώνου μέρος ἤτοι τὸ ΖΘΓ ἢ τὸ ΗΒΚ ἐν τῷ ὑποκειμένῳ, τὸ δὲ λοιπὸν ἐν ἄλλῳ, ἔσται καὶ μιᾶς τῶν ΕΓ, ΕΒ εὐθειῶν μέρος μέν τι ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ, τὸ δὲ ἐν ἄλλῳ.
For if a part of the triangle EGB, namely either ZQG or HBK, is in the underlying plane, and the rest in another, there will also be, of one of the straight lines EG, EB, a part in the underlying plane, and another part in another.
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.
But if the part ZGBH of the triangle EGB be in the underlying plane, and the rest in another, there will also be, of both the straight lines EG, EB, a part in the underlying plane, and another part in another; which has been proved absurd.
τὸ ἄρα ΕΓΒ τρίγωνον ἐν ἑνί ἐστιν ἐπιπέδῳ.
Therefore the triangle EGB is in one plane.
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.
And in that in which the triangle EGB is, in that also is each of EG, EB; and in that in which each of EG, EB is, in that also are AB, GD. Therefore the straight lines AB, GD are in one plane, and every triangle is in one plane; which was to be proved.
§11.prop.3ἐὰν δύο ἐπίπεδα τέμνῃ ἄλληλα, ἡ κοινὴ αὐτῶν τομὴ εὐθεῖά ἐστιν.
If two planes cut one another, their common section is a straight line.
δύο γὰρ ἐπίπεδα τὰ ΑΒ, ΒΓ τεμνέτω ἄλληλα, κοινὴ δὲ αὐτῶν τομὴ ἔστω ἡ ΔΒ γραμμή· λέγω, ὅτι ἡ ΔΒ γραμμὴ εὐθεῖά ἐστιν.
For let two planes AB, BG cut one another, and let their common section be the line DB; I say that the line DB is a straight line.
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.
For, if not, let there be joined from D to B in the plane AB the straight line DEB, and in the plane BG the straight line DZB. Then the two straight lines DEB, DZB will have the same extremities, and will indeed contain a space; which is absurd.
οὐκ ἄρα αἱ ΔΕΒ, ΔΖΒ εὐθεῖαί εἰσιν.
Therefore DEB, DZB are not straight lines.
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.
Similarly indeed we shall show that there will be no other straight line joined from D to B except DB, the common section of the planes AB, BG.
ἐὰν ἄρα δύο ἐπίπεδα τέμνῃ ἄλληλα, ἡ κοινὴ αὐτῶν τομὴ εὐθεῖά ἐστιν· ὅπερ ἔδει δεῖξαι.
Therefore, if two planes cut one another, their common section is a straight line; which was to be proved.