§11.prop.18ἐὰν εὐθεῖα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ πάντα τὰ διʼ αὐτῆς ἐπίπεδα τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται.
If a straight line be perpendicular to any plane, all the planes through it will also be perpendicular to the same plane.
εὐθεῖα γάρ τις ἡ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστω· λέγω, ὅτι καὶ πάντα τὰ διὰ τῆς ΑΒ ἐπίπεδα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
For let any straight line AB be perpendicular to the underlying plane; I say that also all the planes through AB are perpendicular to the underlying plane.
Ἐκβεβλήσθω γὰρ διὰ τῆς ΑΒ ἐπίπεδον τὸ ΔΕ, καὶ ἔστω κοινὴ τομὴ τοῦ ΔΕ ἐπιπέδου καὶ τοῦ ὑποκειμένου ἡ ΓΕ, καὶ εἰλήφθω ἐπὶ τῆς ΓΕ τυχὸν σημεῖον τὸ Ζ, καὶ ἀπὸ τοῦ Ζ τῇ ΓΕ πρὸς ὀρθὰς ἤχθω ἐν τῷ ΔΕ ἐπιπέδῳ ἡ ΖΗ. καὶ ἐπεὶ ἡ ΑΒ πρὸς τὸ ὑποκείμενον ἐπίπεδον ὀρθή ἐστιν, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθή ἐστιν ἡ ΑΒ· ὥστε καὶ πρὸς τὴν ΓΕ ὀρθή ἐστιν· ἡ ἄρα ὑπὸ ΑΒΖ γωνία ὀρθή ἐστιν.
For let a plane DE be drawn through AB, and let the common section of the plane DE and the underlying plane be GE, and let any point Z be taken on GE, and from Z perpendicular to GE let ZH be drawn in the plane DE. And since AB is perpendicular to the underlying plane, therefore also to all the straight lines meeting it and being in the underlying plane, AB is perpendicular; so that it is also perpendicular to GE; therefore the angle ABZ is right.
ἔστι δὲ καὶ ἡ ὑπὸ ΗΖΒ ὀρθή· παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΖΗ. ἡ δὲ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν· καὶ ἡ ΖΗ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
And the angle HZB is also right; therefore AB is parallel to ZH. But AB is perpendicular to the underlying plane; therefore ZH is also to the underlying plane perpendicular.
καὶ ἐπίπεδον πρὸς ἐπίπεδον ὀρθόν ἐστιν, ὅταν αἱ τῇ κοινῇ τομῇ τῶν ἐπιπέδων πρὸς ὀρθὰς ἀγόμεναι εὐθεῖαι ἐν ἑνὶ τῶν ἐπιπέδων τῷ λοιπῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν.
And a plane is perpendicular to a plane when the straight lines drawn perpendicular to the common section of the planes in one of the planes are perpendicular to the remaining plane.
καὶ τῇ κοινῇ τομῇ τῶν ἐπιπέδων τῇ ΓΕ ἐν ἑνὶ τῶν ἐπιπέδων τῷ ΔΕ πρὸς ὀρθὰς ἀχθεῖσα ἡ ΖΗ ἐδείχθη τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς· τὸ ἄρα ΔΕ ἐπίπεδον ὀρθόν ἐστι πρὸς τὸ ὑποκείμενον.
And ZH, drawn perpendicular to the common section of the planes GE in one of the planes DE, was shown perpendicular to the underlying plane; therefore the plane DE is perpendicular to the underlying plane.
ὁμοίως δὴ δειχθήσεται καὶ πάντα τὰ διὰ τῆς ΑΒ ἐπίπεδα ὀρθὰ τυγχάνοντα πρὸς τὸ ὑποκείμενον ἐπίπεδον.
Similarly indeed it will be shown that also all the planes through AB are perpendicular to the underlying plane.
ἐὰν ἄρα εὐθεῖα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ πάντα τὰ διʼ αὐτῆς ἐπίπεδα τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ὅπερ ἔδει δεῖξαι.
Therefore, if a straight line be perpendicular to any plane, all the planes through it will also be perpendicular to the same plane; which was to be proved.
§11.prop.19ἐὰν δύο ἐπίπεδα τέμνοντα ἄλληλα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ κοινὴ αὐτῶν τομὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται.
If two intersecting planes be perpendicular to any plane, their common section will also be perpendicular to the same plane.
δύο γὰρ ἐπίπεδα τὰ ΑΒ, ΒΓ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστω, κοινὴ δὲ αὐτῶν τομὴ ἔστω ἡ ΒΔ· λέγω, ὅτι ἡ ΒΔ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
For let two planes AB, BC be perpendicular to the underlying plane, and let their common section be BD; I say that BD is perpendicular to the underlying plane.
μὴ γάρ, καὶ ἤχθωσαν ἀπὸ τοῦ Δ σημείου ἐν μὲν τῷ ΑΒ ἐπιπέδῳ τῇ ΑΔ εὐθείᾳ πρὸς ὀρθὰς ἡ ΔΕ, ἐν δὲ τῷ ΒΓ ἐπιπέδῳ τῇ ΓΔ πρὸς ὀρθὰς ἡ ΔΖ. καὶ ἐπεὶ τὸ ΑΒ ἐπίπεδον ὀρθόν ἐστι πρὸς τὸ ὑποκείμενον, καὶ τῇ κοινῇ αὐτῶν τομῇ τῇ ΑΔ πρὸς ὀρθὰς ἐν τῷ ΑΒ ἐπιπέδῳ ἦκται ἡ ΔΕ, ἡ ΔΕ ἄρα ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον.
For if not, let there be drawn from the point D, in the plane AB, DE perpendicular to the straight line AD, and in the plane BC, DF perpendicular to CD. And since the plane AB is perpendicular to the underlying plane, and to their common section AD, DE has been drawn perpendicular in the plane AB, therefore DE is perpendicular to the underlying plane.
ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΔΖ ὀρθή ἐστι πρὸς τὸ ὑποκείμενον ἐπίπεδον.
Similarly indeed we shall show that DF is also perpendicular to the underlying plane.
ἀπὸ τοῦ αὐτοῦ ἄρα σημείου τοῦ Δ τῷ ὑποκειμένῳ ἐπιπέδῳ δύο εὐθεῖαι πρὸς ὀρθὰς ἀνεσταμέναι εἰσὶν ἐπὶ τὰ αὐτὰ μέρη· ὅπερ ἐστὶν ἀδύνατον.
Therefore, from the same point D, two straight lines are set up perpendicular to the underlying plane on the same side; which is impossible.
οὐκ ἄρα τῷ ὑποκειμένῳ ἐπιπέδῳ ἀπὸ τοῦ Δ σημείου ἀνασταθήσεται πρὸς ὀρθὰς πλὴν τῆς ΔΒ κοινῆς τομῆς τῶν ΑΒ, ΒΓ ἐπιπέδων.
Therefore, to the underlying plane from the point D, no straight line will be set up perpendicular except the common section DB of the planes AB, BC.
ἐὰν ἄρα δύο ἐπίπεδα τέμνοντα ἄλληλα ἐπιπέδῳ τινὶ πρὸς ὀρθὰς ᾖ, καὶ ἡ κοινὴ αὐτῶν τομὴ τῷ αὐτῷ ἐπιπέδῳ πρὸς ὀρθὰς ἔσται· ὅπερ ἔδει δεῖξαι.
Therefore, if two intersecting planes be perpendicular to any plane, their common section will also be perpendicular to the same plane; which was to be proved.