Measurement of figures: Book 12 Proposition 4
Translations
Ἐὰν ὦσι δύο πυραμίδες ὑπὸ τὸ αὐτὸ ὕψος τριγώνους ἔχουσαι βάσεις, διαιρεθῇ δὲ ἑκατέρα αὐτῶν εἴς τε δύο πυραμίδας ἴσας ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα, ἔσται ὡς ἡ τῆς μιᾶς πυραμίδος βάσις πρὸς τὴν τῆς ἑτέρας πυραμίδος βάσιν, οὕτως τὰ ἐν τῇ μιᾷ πυραμίδι πρίσματα πάντα πρὸς τὰ ἐν τῇ ἑτέρᾳ πυραμίδι πρίσματα πάντα ἰσοπληθῆ. Ἔστωσαν δύο πυραμίδες ὑπὸ τὸ αὐτὸ ὕψος τριγώνους ἔχουσαι βάσεις τὰς ΑΒΓ, ΔΕΖ, κορυφὰς δὲ τὰ Η, Θ σημεῖα, καὶ διῃρήσθω ἑκατέρα αὐτῶν εἴς τε δύο πυραμίδας ἴσας ἀλλήλαις καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα: λέγω, ὅτι ἐστὶν ὡς ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ ἐν τῇ ΑΒΓΗ πυραμίδι πρίσματα πάντα πρὸς τὰ ἐν τῇ ΔΕΖΘ πυραμίδι πρίσματα ἰσοπληθῆ. Ἐπεὶ γὰρ ἴση ἐστὶν ἡ μὲν ΒΞ τῇ ΞΓ, ἡ δὲ ΑΛ τῇ ΛΓ, παράλληλος ἄρα ἐστὶν ἡ ΛΞ τῇ ΑΒ καὶ ὅμοιον τὸ ΑΒΓ τρίγωνον τῷ ΛΞΓ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΔΕΖ τρίγωνον τῷ ΡΦΖ τριγώνῳ ὅμοιόν ἐστιν. καὶ ἐπεὶ διπλασίων ἐστὶν ἡ μὲν ΒΓ τῆς ΓΞ, ἡ δὲ ΕΖ τῆς ΖΦ, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΞ, οὕτως ἡ ΕΖ πρὸς τὴν ΖΦ. καὶ ἀναγέγραπται ἀπὸ μὲν τῶν ΒΓ, ΓΞ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΑΒΓ, ΛΞΓ, ἀπὸ δὲ τῶν ΕΖ, ΖΦ ὅμοιά τε καὶ ὁμοίως κείμενα [εὐθύγραμμα] τὰ ΔΕΖ, ΡΦΖ. ἔστιν ἄρα ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΛΞΓ τρίγωνον, οὕτως τὸ ΔΕΖ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον: ἐναλλὰξ ἄρα ἐστὶν ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ [τρίγωνον], οὕτως τὸ ΛΞΓ [τρίγωνον] πρὸς τὸ ΡΦΖ τρίγωνον. ἀλλ' ὡς τὸ ΛΞΓ τρίγωνον πρὸς τὸ ΡΦΖ τρίγωνον, οὕτως τὸ πρίσμα, οὗ βάσις μὲν [ἐστι] τὸ ΛΞΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΡΦΖ τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ: καὶ ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον, οὕτως τὸ πρίσμα, οὗ βάσις μὲν τὸ ΛΞΓ τρίγωνον, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΡΦΖ τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ. ὡς δὲ τὰ εἰρημένα πρίσματα πρὸς ἄλληλα, οὕτως τὸ πρίσμα, οὗ βάσις μὲν τὸ ΚΒΞΛ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΟΜ εὐθεῖα, πρὸς τὸ πρίσμα, οὗ βάσις μὲν τὸ ΠΕΦΡ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΣΤ εὐθεῖα. καὶ τὰ δύο ἄρα πρίσματα, οὗ τε βάσις μὲν τὸ ΚΒΞΛ παραλληλόγραμμον, ἀπεναντίον δὲ ἡ ΟΜ, καὶ οὗ βάσις μὲν τὸ ΛΞΓ, ἀπεναντίον δὲ τὸ ΟΜΝ, πρὸς τὰ πρίσματα, οὗ τε βάσις μὲν τὸ ΠΕΦΡ, ἀπεναντίον δὲ ἡ ΣΤ εὐθεῖα, καὶ οὗ βάσις μὲν τὸ ΡΦΖ τρίγωνον, ἀπεναντίον δὲ τὸ ΣΤΥ. καὶ ὡς ἄρα ἡ ΑΒΓ βάσις πρὸς τὴν ΔΕΖ βάσιν, οὕτως τὰ εἰρημένα δύο πρίσματα πρὸς τὰ εἰρημένα δύο πρίσματα. Καὶ 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. ἀλλ' 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. 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.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 [τρίγωνον], ἀπεναντίον δὲ τὸ ΣΤΥ, οὕτω δεικτέον. Ἐπὶ γὰρ τῆς αὐτῆς καταγραφῆς νενοήσθωσαν ἀπὸ τῶν Η, Θ κάθετοι ἐπὶ τὰ ΑΒΓ, ΔΕΖ ἐπίπεδα, ἴσαι δηλαδὴ τυγχάνουσαι διὰ τὸ ἰσουψεῖς ὑποκεῖσθαι τὰς πυραμίδας. 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. 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. διὰ τὰ αὐτὰ δὴ καὶ ἡ ἀπὸ τοῦ Θ κάθετος ἐπὶ τὸ ΔΕΖ ἐπίπεδον δίχα τμηθήσεται ὑπὸ τοῦ ΣΤΥ ἐπιπέδου. 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. ἰσουψῆ ἄρα [ἐστὶ] 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 ἰσουψῆ καὶ πρὸς ἄλληλα [εἰσὶν] 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.
If there be two pyramids of the same height which have triangular bases, and cach of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms, then, as the base of the one pyramid is to the base of the other pyramid, so will all the prisms in the one pyramid be to all the prisms, being equal in multitude, in the other pyramid. Let there be two pyramids of the same height which have the triangular bases ABC, DEF, and vertices the points G, H, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; [XII. 3] I say that, as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH, For, since BO is equal to OC, and AL to LC, therefore LO is parallel to AB, and the triangle ABC is similar to the triangle LOC. For the same reason the triangle DEF is also similar to the triangle RVF. And, since BC is double of CO, and EF of FV, therefore, as BC is to CO, so is EF to FV. And on BC, CO are described the similar and similarly situated rectilineal figures ABC, LOC, and on EF, FV the similar and similarly situated figures DEF, RVF; therefore, as the triangle ABC is to the triangle LOC, so is the triangle DEF to the triangle RVF; [VI. 22] therefore, alternately, as the triangle ABC is to the triangle DEF, so is the triangle LOC to the triangle RVF. [V. 16] But, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite; [Lemma following] therefore also, as the triangle ABC is to the triangle DEF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite. But, as the said prisms are to one another, so is the prism in which the parallelogram KBOL is the base and the straight line PM its opposite to the prism in which the parallelogram QEVR is the base and the straight line ST its opposite. [XI. 39XII. 3] Therefore also the two prisms, that in which the parallelogram KBOL is the base and PM its opposite, and that in which the triangle LOC is the base and PMN its opposite, are to the prisms in which QEVR is the base and the straight line ST its opposite and in which the triangle RVF is the base and STU its opposite in the same ratio [V. 12] Therefore also, as the base ABC is to the base DEF, so are the said two prisms to the said two prisms. And similarly, if the pyramids PMNG, STUH be divided into two prisms and two pyramids, as the base PMN is to the base STU, so will the two prisms in the pyramid PMNG be to the two prisms in the pyramid STUH. But, as the base PMN is to the base STU, so is the base ABC to the base DEF; for the triangles PMN, STU are equal to the triangles LOC, RVF respectively. Therefore also, as the base ABC is to the base DEF, so are the four prisms to the four prisms. And similarly also, if we divide the remaining pyramids into two pyramids and into two prisms, then, as the base ABC is to the base DEF, so will all the prisms in the pyramid ABCG be to all the prisms, being equal in multitude, in the pyramid DEFH. Q. E. D.Lemma. But that, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite, we must prove as follows. For in the same figure let perpendiculars be conceived drawn from G, H to the planes ABC, DEF; these are of course equal because, by hypothesis, the pyramids are of equal height. Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC, PMN, they will be cut in the same ratios. [XI. 17] And GC is bisected by the plane PMN at N; therefore the perpendicular from G to the plane ABC will also be bisected by the plane PMN. For the same reason the perpendicular from H to the plane DEF will also be bisected by the plane STU. And the perpendiculars from G, H to the planes ABC, DEF are equal; therefore the perpendiculars from the triangles PMN, STU to the planes ABC, DEF are also equal. Therefore the prisms in which the triangles LOC, RVF are bases, and PMN, STU their opposites, are of equal height.