Solid geometry: Book 11 Proposition 28
Translations
Ἐὰν στερεὸν παραλληλεπίπεδον ἐπιπέδῳ τμηθῇ κατὰ τὰς διαγωνίους τῶν ἀπεναντίον ἐπιπέδων, δίχα τμηθήσεται τὸ στερεὸν ὑπὸ τοῦ ἐπιπέδου. Στερεὸν γὰρ παραλληλεπίπεδον τὸ ΑΒ ἐπιπέδῳ τῷ ΓΔΕΖ τετμήσθω κατὰ τὰς διαγωνίους τῶν ἀπεναντίον ἐπιπέδων τὰς ΓΖ, ΔΕ: λέγω, ὅτι δίχα τμηθήσεται τὸ ΑΒ στερεὸν ὑπὸ τοῦ ΓΔΕΖ ἐπιπέδου. Ἐπεὶ γὰρ ἴσον ἐστὶ τὸ μὲν ΓΗΖ τρίγωνον τῷ ΓΖΒ τριγώνῳ, τὸ δὲ ΑΔΕ τῷ ΔΕΘ, ἔστι δὲ καὶ τὸ μὲν ΓΑ παραλληλόγραμμον τῷ ΕΒ ἴσον: ἀπεναντίον γάρ: τὸ δὲ ΗΕ τῷ ΓΘ, καὶ τὸ πρίσμα ἄρα τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων τῶν ΓΗΖ, ΑΔΕ, τριῶν δὲ παραλληλογράμμων τῶν ΗΕ, ΑΓ, ΓΕ ἴσον ἐστὶ τῷ πρίσματι τῷ περιεχομένῳ ὑπὸ δύο μὲν τριγώνων τῶν ΓΖΒ, ΔΕΘ, τριῶν δὲ παραλληλογράμμων τῶν ΓΘ, ΒΕ, ΓΕ: ὑπὸ γὰρ ἴσων ἐπιπέδων περιέχονται τῷ τε πλήθει καὶ τῷ μεγέθει. ὥστε ὅλον τὸ ΑΒ στερεὸν δίχα τέτμηται ὑπὸ τοῦ ΓΔΕΖ ἐπιπέδου: ὅπερ ἔδει δεῖξαι.
If a parallelepipedal solid be cut by a plane through the diagonals of the opposite planes, the solid will be bisected by the plane. For let the parallelepipedal solid AB be cut by the plane CDEF through the diagonals CF, DE of opposite planes; I say that the solid AB will be bisected by the plane CDEF. For, since the triangle CGF is equal to the triangle CFB, [I. 34] and ADE to DEH, while the parallelogram CA is also equal to the parallelogram EB, for they are opposite, and GE to CH, therefore the prism contained by the two triangles CGF, ADE and the three parallelograms GE, AC, CE is also equal to the prism contained by the two triangles CFB, DEH and the three parallelograms CH, BE, CE; for they are contained by planes equal both in multitude and in magnitude. [XI. Def. 10]