§11.prop.34#2ἀλλʼ ὡς μὲν ἡ ΕΘ πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν· ἰσοϋψῆ γάρ ἐστι τὰ ΑΒ, ΓΦ στερεά· ὡς δὲ ἡ ΓΜ πρὸς τὴν ΓΤ, οὕτως ἥ τε ΜΠ βάσις πρὸς τὴν ΠΤ βάσιν καὶ τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν.
But as the base EQ is to the base NP, so is the solid AB to the solid GF; for the solids AB, GF are of the same height; and as GM is to GT, so is both the base MP to the base PT, and the solid GD to the solid GF.
καὶ ὡς ἄρα τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν· ἑκάτερον ἄρα τῶν ΑΒ, ΓΔ πρὸς τὸ ΓΦ τὸν αὐτὸν ἔχει λόγον.
And as therefore the solid AB is to the solid GF, so is the solid GD to the solid GF; therefore each of the solids AB, GD has to the solid GF the same ratio.
ἴσον ἄρα ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ.
Therefore the solid AB is equal to the solid GD.
μὴ ἔστωσαν δὴ αἱ ἐφεστηκυῖαι αἱ ΖΕ, ΒΛ, ΗΑ, ΘΚ, ΞΝ, ΔΟ, ΜΓ, ΡΠ πρὸς ὀρθὰς ταῖς βάσεσιν αὐτῶν, καὶ ἤχθωσαν ἀπὸ τῶν Ζ, Η, Β, Κ, Ξ, Μ, Δ, Ρ σημείων ἐπὶ τὰ διὰ τῶν ΕΘ, ΝΠ ἐπίπεδα κάθετοι καὶ συμβαλλέτωσαν τοῖς ἐπιπέδοις κατὰ τὰ Σ, Τ, Υ, Φ, Χ, Ψ, Ω, #2, καὶ συμπεπληρώσθω τὰ ΖΦ, ΞΩ στερεά· λέγω, ὅτι καὶ οὕτως ἴσων ὄντων τῶν ΑΒ, ΓΔ στερεῶν ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, καί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος.
Now let the standing straight lines ZE, BL, HA, QK, XN, DO, MG, RP not be at right angles to their bases, and from the points Z, H, B, K, X, M, D, R let perpendiculars be drawn to the planes through EQ, NP, and let them meet the planes at the points S, T, Y, F, Ch, Ps, O, #2, and let the solids ZF, XW be completed; I say that, even so, if the solids AB, GD are equal, their bases are reciprocally proportional to their heights, and as the base EQ is to the base NP, so is the height of the solid GD to the height of the solid AB.
ἐπεὶ ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ, ἀλλὰ τὸ μὲν ΑΒ τῷ ΒΤ ἐστιν ἴσον· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΖΚ καὶ ὑπὸ τὸ αὐτὸ ὕψος· τὸ δὲ ΓΔ στερεὸν τῷ ΔΨ ἐστιν ἴσον· ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς ΡΞ καὶ ὑπὸ τὸ αὐτὸ ὕψος· καὶ τὸ ΒΤ ἄρα στερεὸν τῷ ΔΨ στερεῷ ἴσον ἐστίν.
For since the solid AB is equal to the solid GD, but the solid AB is equal to the solid BT; for they are on the same base ZK and of the same height; and the solid GD is equal to the solid DY; for they are again on the same base RX and of the same height; therefore the solid BT is also equal to the solid DY.
ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ στερεοῦ ὕψος.
Therefore as the base ZK is to the base XR, so is the height of the solid DY to the height of the solid BT.
ἴση δὲ ἡ μὲν ΖΚ βάσις τῇ ΕΘ βάσει, ἡ δὲ ΞΡ βάσις τῇ ΝΠ βάσει· ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ στερεοῦ ὕψος.
But the base ZK is equal to the base EQ, and the base XR to the base NP; therefore as the base EQ is to the base NP, so is the height of the solid DY to the height of the solid BT.
τὰ δʼ αὐτὰ ὕψη ἐστὶ τῶν ΔΨ, ΒΤ στερεῶν καὶ τῶν ΔΓ, ΒΑ· ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΔΓ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος.
And the heights of the solids DY, BT are the same as those of the solids DG, BA; therefore as the base EQ is to the base NP, so is the height of the solid DG to the height of the solid AB.
τῶν ΑΒ, ΓΔ ἄρα στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν.
Therefore of the solid parallelepipeds AB, GD the bases are reciprocally proportional to their heights.
πάλιν δὴ τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντιπεπονθέτωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ.
Again, let the bases of the solid parallelepipeds AB, GD be reciprocally proportional to their heights, and let the base EQ be to the base NP as the height of the solid GD is to the height of the solid AB; I say that the solid AB is equal to the solid GD.
τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος, ἴση δὲ ἡ μὲν ΕΘ βάσις τῇ ΖΚ βάσει, ἡ δὲ ΝΠ τῇ ΞΡ, ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος.
For with the same constructions made, since as the base EQ is to the base NP, so is the height of the solid GD to the height of the solid AB, and the base EQ is equal to the base ZK, and the base NP to the base XR, therefore as the base ZK is to the base XR, so is the height of the solid GD to the height of the solid AB.
τὰ δʼ αὐτὰ ὕψη ἐστὶ τῶν ΑΒ, ΓΔ στερεῶν καὶ τῶν ΒΤ, ΔΨ· ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ στερεοῦ ὕψος.
But the heights of the solids AB, GD are the same as those of the solids BT, DY; therefore as the base ZK is to the base XR, so is the height of the solid DY to the height of the solid BT.
τῶν ΒΤ, ΔΨ ἄρα στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν· ἴσον ἄρα ἐστὶ τὸ ΒΤ στερεὸν τῷ ΔΨ στερεῷ.
Therefore of the solid parallelepipeds BT, DY the bases are reciprocally proportional to their heights; therefore the solid BT is equal to the solid DY.
ἀλλὰ τὸ μὲν ΒΤ τῷ ΒΑ ἴσον ἐστίν· ἐπί τε γὰρ τῆς αὐτῆς βάσεως τῆς ΖΚ καὶ ὑπὸ τὸ αὐτὸ ὕψος. τὸ δὲ ΔΨ στερεὸν τῷ ΔΓ στερεῷ ἴσον ἐστίν. καὶ τὸ ΑΒ ἄρα στερεὸν τῷ ΓΔ στερεῷ ἐστιν ἴσον· ὅπερ ἔδει δεῖξαι.
But the solid BT is equal to the solid BA; for they are on the same base ZK and of the same height; and the solid DY is equal to the solid DG; therefore the solid AB is also equal to the solid GD; which was to be proved.