§11.prop.16ἐὰν δύο ἐπίπεδα παράλληλα ὑπὸ ἐπιπέδου τινὸς τέμνηται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν.
If two parallel planes be cut by any plane, their common sections are parallel.
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.
For let two parallel planes AB, CD be cut by the plane EFGH, and let their common sections be EF, GH; I say that EF is parallel to GH. For if not, EF, GH being produced will meet either on the side of F, H or on the side of E, G.
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.
Let them be produced on the side of F, H and let them meet first at K. And since EFK is in the plane AB, therefore also all points on EFK are in the plane AB.
ἓν δὲ τῶν ἐπὶ τῆς ΕΖΚ εὐθείας σημείων ἐστὶ τὸ Κ· τὸ Κ ἄρα ἐν τῷ ΑΒ ἐστιν ἐπιπέδῳ.
And K is one of the points on the straight line EFK; therefore K is in the plane AB.
διὰ τὰ αὐτὰ δὴ τὸ Κ καὶ ἐν τῷ ΓΔ ἐστιν ἐπιπέδῳ· τὰ ΑΒ, ΓΔ ἄρα ἐπίπεδα ἐκβαλλόμενα συμπεσοῦνται.
For the same reasons indeed, K is also in the plane CD; therefore the planes AB, CD being produced will meet.
οὐ συμπίπτουσι δὲ διὰ τὸ παράλληλα ὑποκεῖσθαι· οὐκ ἄρα αἱ ΕΖ, ΗΘ εὐθεῖαι ἐκβαλλόμεναι ἐπὶ τὰ Ζ, Θ μέρη συμπεσοῦνται.
But they do not meet, because they are assumed parallel; therefore the straight lines EF, GH being produced on the side of F, H will not meet.
ὁμοίως δὴ δείξομεν, ὅτι αἱ ΕΖ, ΗΘ εὐθεῖαι οὐδὲ ἐπὶ τὰ Ε, Η μέρη ἐκβαλλόμεναι συμπεσοῦνται.
Similarly indeed we shall show that the straight lines EF, GH being produced on the side of E, G will not meet either.
αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη συμπίπτουσαι παράλληλοί εἰσιν.
And straight lines which do not meet on either side are parallel.
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.
Therefore EF is parallel to GH. Therefore, if two parallel planes be cut by any plane, their common sections are parallel; which was to be proved.
§11.prop.17ἐὰν δύο εὐθεῖαι ὑπὸ παραλλήλων ἐπιπέδων τέμνωνται εἰς τοὺς αὐτοὺς λόγους τμηθήσονται.
If two straight lines be cut by parallel planes, they will be cut in the same ratios.
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.
For let two straight lines AB, CD be cut by parallel planes HT, KL, MN at the points A, E, B, 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, and let AD meet the plane KL at the point X, and let EX, XF be joined. And since two parallel planes KL, MN are cut by the plane EBDX, their common sections EX, BD are parallel.
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.
For the same reasons indeed, since two parallel planes HT, KL are cut by the plane AXFC, their common sections AC, XF are parallel.
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.
And since in the triangle ABD, EX has been drawn parallel to BD, one of the sides, therefore proportionally, as AE is to EB, so is AX to XD. Again, since in the triangle ADC, XF has been drawn parallel to AC, one of the sides, proportionally, as AX is to XD, so is CF to FD. And it was also shown that, as AX is to XD, so is AE to EB; therefore also, as AE is to EB, so is CF to FD. Therefore, if two straight lines be cut by parallel planes, they will be cut in the same ratios; which was to be proved.