§12.prop.4#2καὶ ὁμοίως, ἐὰν διαιρεθῶσιν αἱ ΟΜΝΗ, ΣΤΥΘ πυραμίδες εἴς τε δύο πρίσματα καὶ δύο πυραμίδας, ἔσται ὡς ἡ ΟΜΝ βάσις πρὸς τὴν ΣΤΥ βάσιν, οὕτως τὰ ἐν τῇ ΟΜ ΝΗ πυραμίδι δύο πρίσματα πρὸς τὰ ἐν τῇ ΣΤΥΘ πυραμίδι δύο πρίσματα.
And similarly, if the pyramids OMNH, STYQ are divided into two prisms and two pyramids, as the base OMN is to the base STY, so will the two prisms in the pyramid OMNH be to the two prisms in the pyramid STYQ.
ἀλλʼ ὡς ἡ ΟΜΝ βάσις πρὸς τὴν ΣΤΥ βάσιν, οὕτως ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν· ἴσον γὰρ ἑκάτερον τῶν ΟΜΝ, ΣΤΥ τριγώνων ἑκατέρῳ τῶν ΛΞΓ, ΡΦΖ. καὶ ὡς ἄρα ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ τέσσαρα πρίσματα πρὸς τὰ τέσσαρα πρίσματα.
But as the base OMN is to the base STY, so is the base ABG to the base DEZ; for each of the triangles OMN, STY is equal to each of the triangles LXG, RFZ. And therefore, as the base ABG is to the base DEZ, so are the four prisms to the four prisms.
ὁμοίως δὲ κἂν τὰς ὑπολειπομένας πυραμίδας διέλωμεν εἴς τε δύο πυραμίδας καὶ εἰς δύο πρίσματα, ἔσται ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ ἐν τῇ ΑΒ ΓΗ πυραμίδι πρίσματα πάντα πρὸς τὰ ἐν τῇ ΔΕΖΘ πυραμίδι πρίσματα πάντα ἰσοπληθῆ· ὅπερ ἔδει δεῖξαι.
And similarly, even if we divide the remaining pyramids into two pyramids and into two prisms, as the base ABG is to the base DEZ, so will all the prisms in the pyramid ABGH be to all the prisms, equal in number, in the pyramid DEZQ; which was to be proved.
λῆμμα
ὅτι δέ ἐστιν ὡς τὸ ΛΞΓ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον, οὕτως τὸ πρίσμα, οὗ βάσις τὸ ΛΞΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΡΦΖ, ἀπεναντίον δὲ τὸ ΣΤΥ, οὕτω δεικτέον.
Lemma That, as the triangle LXG is to the triangle RFZ, so is the prism of which the base is the triangle LXG, and the opposite the triangle OMN, to the prism of which the base is RFZ, and the opposite the triangle STY, must be proved as follows.
ἐπὶ γὰρ τῆς αὐτῆς καταγραφῆς νενοήσθωσαν ἀπὸ τῶν Η, Θ κάθετοι ἐπὶ τὰ ΑΒΓ, ΔΕΖ ἐπίπεδα, ἴσαι δηλαδὴ τυγχάνουσαι διὰ τὸ ἰσοϋψεῖς ὑποκεῖσθαι τὰς πυραμίδας.
For on the same figure, let there be conceived perpendiculars from the points H, Q to the planes ABG, DEZ, which are of course equal because the pyramids are assumed to be of equal height.
καὶ ἐπεὶ δύο εὐθεῖαι ἥ τε ΗΓ καὶ ἡ ἀπὸ τοῦ Η κάθετος ὑπὸ παραλλήλων ἐπιπέδων τῶν ΑΒΓ, ΟΜΝ τέμνονται, εἰς τοὺς αὐτοὺς λόγους τμηθήσονται.
And since two straight lines, namely HG and the perpendicular from H, are cut by the parallel planes ABG, OMN, they will be cut in the same ratios.
καὶ τέτμηται ἡ ΗΓ δίχα ὑπὸ τοῦ ΟΜΝ ἐπιπέδου κατὰ τὸ Ν· καὶ ἡ ἀπὸ τοῦ Η ἄρα κάθετος ἐπὶ τὸ ΑΒΓ ἐπίπεδον δίχα τμηθήσεται ὑπὸ τοῦ ΟΜΝ ἐπιπέδου.
And HG has been cut in half by the plane OMN at N; therefore the perpendicular from H to the plane ABG will also be cut in half by the plane OMN.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ἀπὸ τοῦ Θ κάθετος ἐπὶ τὸ ΔΕΖ ἐπίπεδον δίχα τμηθήσεται ὑπὸ τοῦ ΣΤΥ ἐπιπέδου.
For the same reason, the perpendicular from Q to the plane DEZ will also be cut in half by the plane STY.
καί εἰσιν ἴσαι αἱ ἀπὸ τῶν Η, Θ κάθετοι ἐπὶ τὰ ΑΒΓ, ΔΕΖ ἐπίπεδα· ἴσαι ἄρα καὶ αἱ ἀπὸ τῶν ΟΜΝ, ΣΤΥ τριγώνων ἐπὶ τὰ ΑΒΓ, ΔΕΖ κάθετοι.
And the perpendiculars from the points H, Q to the planes ABG, DEZ are equal; therefore the perpendiculars from the triangles OMN, STY to the planes ABG, DEZ are also equal.
ἰσοϋψῆ ἄρα τὰ πρίσματα, ὧν βάσεις μέν εἰσι τὰ ΛΞΓ, ΡΦΖ τρίγωνα, ἀπεναντίον δὲ τὰ ΟΜΝ, ΣΤΥ. ὥστε καὶ τὰ στερεὰ παραλληλεπίπεδα τὰ ἀπὸ τῶν εἰρημένων πρισμάτων ἀναγραφόμενα ἰσοϋψῆ καὶ πρὸς ἄλληλα ὡς αἱ βάσεις· καὶ τὰ ἡμίση ἄρα ἐστὶν ὡς ἡ ΛΞΓ βάσις πρὸς τὴν ΡΦΖ βάσιν, οὕτως τὰ εἰρημένα πρίσματα πρὸς ἄλληλα· ὅπερ ἔδει δεῖξαι.
Therefore, the prisms of which the bases are the triangles LXG, RFZ, and the opposites OMN, STY, are of equal height. Therefore also the solid parallelepipeds described from the aforesaid prisms are of equal height and have to one another the same ratio as their bases; therefore their halves also, as the base LXG is to the base RFZ, so are the aforesaid prisms to one another; which was to be proved.