28. 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, [] 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. [] Hence the whole solid AB is bisected by the plane CDEF. Q. E. D.
28. 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.
29. Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are on the same straight lines, are equal to one another. Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AG, AF, LM, LN, CD, CE, BH, BK, be on the same straight lines FN, DK; I say that the solid CM is equal to the solid CN. For, since each of the figures CH, CK is a parallelogram, CB is equal to each of the straight lines DH, EK, [] hence DH is also equal to EK. Let EH be subtracted from each; therefore the remainder DE is equal to the remainder HK. Hence the triangle DCE is also equal to the triangle HBK, [] and the parallelogram DG to the parallelogram HN. [] For the same reason the triangle AFG is also equal to the triangle MLN. But the parallelogram CF is equal to the parallelogram BM, and CG to BN, for they are opposite; therefore the prism contained by the two triangles AFG, DCE and the three parallelograms AD, DG, CG is equal to the prism contained by the two triangles MLN, HBK and the three parallelograms BM, HN, BN. Let there be added to each the solid of which the parallelogram AB is the base and GEHM its opposite; therefore the whole parallelepipedal solid CM is equal to the whole parallelepipedal solid CN. Therefore etc. Q. E. D.
29. τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν. ἔστω ἐπὶ τῆς αὐτῆς βάσεως τῆς ΑΒ στερεὰ παραλληλεπίπεδα τὰ ΓΜ, ΓΝ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΗ, ΑΖ, ΛΜ, ΛΝ, ΓΔ, ΓΕ, ΒΘ, ΒΚ ἐπὶ τῶν αὐτῶν εὐθειῶν ἔστωσαν τῶν ΖΝ, ΔΚ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΓΜ στερεὸν τῷ ΓΝ στερεῷ. ἐπεὶ γὰρ παραλληλόγραμμόν ἐστιν ἑκάτερον τῶν ΓΘ, ΓΚ, ἴση ἐστὶν ἡ ΓΒ ἑκατέρᾳ τῶν ΔΘ, ΕΚ· ὥστε καὶ ἡ ΔΘ τῇ ΕΚ ἐστιν ἴση. κοινὴ ἀφῃρήσθω ἡ ΕΘ· λοιπὴ ἄρα ἡ ΔΕ λοιπῇ τῇ ΘΚ ἐστιν ἴση. ὥστε καὶ τὸ μὲν ΔΓΕ τρίγωνον τῷ ΘΒΚ τριγώνῳ ἴσον ἐστίν, τὸ δὲ ΔΗ παραλληλόγραμμον τῷ ΘΝ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΖΗ τρίγωνον τῷ ΜΛΝ τριγώνῳ ἴσον ἐστίν. ἔστι δὲ καὶ τὸ μὲν ΓΖ παραλληλόγραμμον τῷ ΒΜ παραλληλογράμμῳ ἴσον, τὸ δὲ ΓΗ τῷ ΒΝ· ἀπεναντίον γάρ· καὶ τὸ πρίσμα ἄρα τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων τῶν ΑΖΗ, ΔΓΕ, τριῶν δὲ παραλληλογράμμων τῶν ΑΔ, ΔΗ, ΓΗ ἴσον ἐστὶ τῷ πρίσματι τῷ περιεχομένῳ ὑπὸ δύο μὲν τριγώνων τῶν ΜΛΝ, ΘΒΚ, τριῶν δὲ παραλληλογράμμων τῶν ΒΜ, ΘΝ, ΒΝ. κοινὸν προσκείσθω τὸ στερεόν, οὗ βάσις μὲν τὸ ΑΒ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΗΕΘΜ· ὅλον ἄρα τὸ ΓΜ στερεὸν παραλληλεπίπεδον ὅλῳ τῷ ΓΝ στερεῷ παραλληλεπιπέδῳ ἴσον ἐστίν. τὰ ἄρα ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
30. Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are not on the same straight lines, are equal to one another. Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AF, AG, LM, LN, CD, CE, BH, BK, not be on the same straight lines; I say that the solid CM is equal to the solid CN. For let NK, DH be produced and meet one another at R, and further let FM, GE be produced to P, Q; let AO, LP, CQ, BR be joined. Then the solid CM, of which the parallelogram ACBL is the base, and FDHM its opposite, is equal to the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite; for they are on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AF, AO, LM, LP, CD, CQ, BH, BR, are on the same straight lines FP, DR. [] But the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite, is equal to the solid CN, of which the parallelogram ACBL is the base and GEKN its opposite; for they are again on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AG, AO, CE, CQ, LN, LP, BK, BR, are on the same straight lines GQ, NR. Hence the solid CM is also equal to the solid CN. Therefore etc. Q. E. D.
30. τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν. ἔστω ἐπὶ τῆς αὐτῆς βάσεως τῆς ΑΒ στερεὰ παραλληλεπίπεδα τὰ ΓΜ, ΓΝ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΖ, ΑΗ, ΛΜ, ΛΝ, ΓΔ, ΓΕ, ΒΘ, ΒΚ μὴ ἔστωσαν ἐπὶ τῶν αὐτῶν εὐθειῶν· λέγω, ὅτι ἴσον ἐστὶ τὸ ΓΜ στερεὸν τῷ ΓΝ στερεῷ. ἐκβεβλήσθωσαν γὰρ αἱ ΝΚ, ΔΘ καὶ συμπιπτέτωσαν ἀλλήλαις κατὰ τὸ Ρ, καὶ ἔτι ἐκβεβλήσθωσαν αἱ ΖΜ, ΗΕ ἐπὶ τὰ Ο, Π, καὶ ἐπεζεύχθωσαν αἱ ΑΞ, ΛΟ, ΓΠ, ΒΡ. ἴσον δή ἐστι τὸ ΓΜ στερεόν, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΖΔΘΜ, τῷ ΓΟ στερεῷ, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΞΠΡΟ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΑΓΒΛ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΖ, ΑΞ, ΛΜ, ΛΟ, ΓΔ, ΓΠ, ΒΘ, ΒΡ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΖΟ, ΔΡ. ἀλλὰ τὸ ΓΟ στερεόν, οὗ βάσις μέν ἐστι τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΞΠΡΟ, ἴσον ἐστὶ τῷ ΓΝ στερεῷ, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΗΕΚΝ· ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς ΑΓΒΛ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΗ, ΑΞ, ΓΕ, ΓΠ, ΛΝ, ΛΟ, ΒΚ, ΒΡ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΗΠ, ΝΡ. ὥστε καὶ τὸ ΓΜ στερεὸν ἴσον ἐστὶ τῷ ΓΝ στερεῷ. τὰ ἄρα ἐπὶ τῆς αὐτῆς βάσεως στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
31. Parallelepipedal solids which are on equal bases and of the same height are equal to one another. Let the parallelepipedal solids AE, CF, of the same height, be on equal bases AB, CD. I say that the solid AE is equal to the solid CF. First, let the sides which stand up, HK, BE, AG, LM, PQ, DF, CO, RS, be at right angles to the bases AB, CD; let the straight line RT be produced in a straight line with CR; on the straight line RT, and at the point R on it, let the angle TRU be constructed equal to the angle ALB, [] let RT be made equal to AL, and RU equal to LB, and let the base RW and the solid XU be completed. Now, since the two sides TR, RU are equal to the two sides AL, LB, and they contain equal angles, therefore the parallelogram RW is equal and similar to the parallelogram HL. Since again AL is equal to RT, and LM to RS, and they contain right angles, therefore the parallelogram RX is equal and similar to the parallelogram AM. For the same reason LE is also equal and similar to SU; therefore three parallelograms of the solid AE are equal and similar to three parallelograms of the solid XU. But the former three are equal and similar to the three opposite, and the latter three to the three opposite; [] therefore the whole parallelepipedal solid AE is equal to the whole parallelepipedal solid XU. [] Let DR, WU be drawn through and meet one another at Y, let aTb be drawn through T parallel to DY, let PD be produced to a, and let the solids YX, RI be completed. Then the solid XY, of which the parallelogram RX is the base and Yc its opposite, is equal to the solid XU of which the parallelogram RX is the base and UV its opposite, for they are on the same base RX and of the same height, and the extremities of their sides which stand up, namely RY, RU, Tb, TW, Se, Sd, Xc, XV, are on the same straight lines YW, eV. [] But the solid XU is equal to AE: therefore the solid XY is also equal to the solid AE. And, since the parallelogram RUWT is equal to the parallelogram YT for they are on the same base RT and in the same parallels RT, YW, [] while RUWT is equal to CD, since it is also equal to AB, therefore the parallelogram YT is also equal to CD. But DT is another parallelogram; therefore, as the base CD is to DT, so is YT to DT. [] And, since the parallelepipedal solid CI has been cut by the plane RF which is parallel to opposite planes, as the base CD is to the base DT, so is the solid CF to the solid RI. [] For the same reason, since the parallelepipedal solid YI has been cut by the plane RX which is parallel to opposite planes, as the base YT is to the base TD, so is the solid YX to the solid RI. [] But, as the base CD is to DT, so is YT to DT; therefore also, as the solid CF is to the solid RI, so is the solid YX to RI. [] Therefore each of the solids CF, YX has to RI the same ratio; therefore the solid CF is equal to the solid YX. [] But YX was proved equal to AE; therefore AE is also equal to CF. Next, let the sides standing up, AG, HK, BE, LM, CN, PQ, DF, RS, not be at right angles to the bases AB, CD; I say again that the solid AE is equal to the solid CF. For from the points K, E, G, M, Q, F, N, S let KO, ET, GU, MV, QW, FX, NY, SI be drawn perpendicular to the plane of reference, and let them meet the plane at the points O, T, U, V, W, X, Y, I, and let OT, OU, UV, TV, WX, WY, YI, IX be joined. Then the solid KV is equal to the solid QI, for they are on the equal bases KM, QS and of the same height, and their sides which stand up are at right angles to their bases. [First part of this Prop.] But the solid KV is equal to the solid AE, and QI to CF; for they are on the same base and of the same height, while the extremities of their sides which stand up are not on the same straight lines. [] Therefore the solid AE is also equal to the solid CF. Therefore etc. Q. E. D.
31. τὰ ἐπὶ ἴσων βάσεων ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν. ἔστω ἐπὶ ἴσων βάσεων τῶν ΑΒ, ΓΔ στερεὰ παραλληλεπίπεδα τὰ ΑΕ, ΓΖ ὑπὸ τὸ αὐτὸ ὕψος· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΕ στερεὸν τῷ ΓΖ στερεῷ. ἔστωσαν δὴ πρότερον αἱ ἐφεστηκυῖαι αἱ ΘΚ, ΒΕ, ΑΗ, ΛΜ, ΟΠ, ΔΖ, ΓΞ, ΡΣ πρὸς ὀρθὰς ταῖς ΑΒ, ΓΔ βάσεσιν, καὶ ἐκβεβλήσθω ἐπʼ εὐθείας τῇ ΓΡ εὐθεῖα ἡ ΡΤ, καὶ συνεστάτω πρὸς τῇ ΡΤ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ρ τῇ ὑπὸ ΑΛΒ γωνίᾳ ἴση ἡ ὑπὸ ΤΡΥ, καὶ κείσθω τῇ μὲν ΑΛ ἴση ἡ ΡΤ, τῇ δὲ ΛΒ ἴση ἡ ΡΥ, καὶ συμπεπληρώσθω ἥ τε ΡΧ βάσις καὶ τὸ ΨΥ στερεόν. καὶ ἐπεὶ δύο αἱ ΤΡ, ΡΥ δυσὶ ταῖς ΑΛ, ΛΒ ἴσαι εἰσίν, καὶ γωνίας ἴσας περιέχουσιν, ἴσον ἄρα καὶ ὅμοιον τὸ ΡΧ παραλληλόγραμμον τῷ ΘΛ παραλληλογράμμῳ. καὶ ἐπεὶ πάλιν ἴση μὲν ἡ ΑΛ τῇ ΡΤ, ἡ δὲ ΛΜ τῇ ΡΣ, καὶ γωνίας ὀρθὰς περιέχουσιν, ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΡΨ παραλληλόγραμμον τῷ ΑΜ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΛΕ τῷ ΣΥ ἴσον τέ ἐστι καὶ ὅμοιον· τρία ἄρα παραλληλόγραμμα τοῦ ΑΕ στερεοῦ τρισὶ παραλληλογράμμοις τοῦ ΨΥ στερεοῦ ἴσα τέ ἐστι καὶ ὅμοια. ἀλλὰ τὰ μὲν τρία τρισὶ τοῖς ἀπεναντίον ἴσα τέ ἐστι καὶ ὅμοια, τὰ δὲ τρία τρισὶ τοῖς ἀπεναντίον· ὅλον ἄρα τὸ ΑΕ στερεὸν παραλληλεπίπεδον ὅλῳ τῷ ΨΥ στερεῷ παραλληλεπιπέδῳ ἴσον ἐστίν. διήχθωσαν αἱ ΔΡ, ΧΥ καὶ συμπιπτέτωσαν ἀλλήλαις κατὰ τὸ Ω, καὶ διὰ τοῦ Τ τῇ ΔΩ παράλληλος ἤχθω ἡ #22αΤ#5, καὶ ἐκβεβλήσθω ἡ ΟΔ κατὰ τὸ #22α, καὶ συμπεπληρώσθω τὰ ΩΨ, ΡΙ στερεά. ἴσον δή ἐστι τὸ ΨΩ στερεόν, οὗ βάσις μέν ἐστι τὸ ΡΨ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ Ω#4, τῷ ΨΥ στερεῷ, οὗ βάσις μὲν τὸ ΡΨ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΥΦ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΡΨ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΡΩ, ΡΥ, Τ#5, ΤΧ, Σ#2, Σο͂, Ψ#4, ΨΦ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΩΧ, #2Φ. ἀλλὰ τὸ ΨΥ στερεὸν τῷ ΑΕ ἐστιν ἴσον· καὶ τὸ ΨΩ ἄρα στερεὸν τῷ ΑΕ στερεῷ ἐστιν ἴσον. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΡΥΧΤ παραλληλόγραμμον τῷ ΩΤ παραλληλογράμμῳ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΡΤ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΡΤ, ΩΧ· ἀλλὰ τὸ ΡΥΧΤ τῷ ΓΔ ἐστιν ἴσον, ἐπεὶ καὶ τῷ ΑΒ, καὶ τὸ ΩΤ ἄρα παραλληλόγραμμον τῷ ΓΔ ἐστιν ἴσον. ἄλλο δὲ τὸ ΔΤ· ἔστιν ἄρα ὡς ἡ ΓΔ βάσις πρὸς τὴν ΔΤ, οὕτως ἡ ΩΤ πρὸς τὴν ΔΤ. καὶ ἐπεὶ στερεὸν παραλληλεπίπεδον τὸ ΓΙ ἐπιπέδῳ τῷ ΡΖ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ὡς ἡ ΓΔ βάσις πρὸς τὴν ΔΤ βάσιν, οὕτως τὸ ΓΖ στερεὸν πρὸς τὸ ΡΙ στερεόν. διὰ τὰ αὐτὰ δή, ἐπεὶ στερεὸν παραλληλεπίπεδον τὸ ΩΙ ἐπιπέδῳ τῷ ΡΨ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ὡς ἡ ΩΤ βάσις πρὸς τὴν ΤΔ βάσιν, οὕτως τὸ ΩΨ στερεὸν πρὸς τὸ ΡΙ. ἀλλʼ ὡς ἡ ΓΔ βάσις πρὸς τὴν ΔΤ, οὕτως ἡ ΩΤ πρὸς τὴν ΔΤ· καὶ ὡς ἄρα τὸ ΓΖ στερεὸν πρὸς τὸ ΡΙ στερεόν, οὕτως τὸ ΩΨ στερεὸν πρὸς τὸ ΡΙ. ἑκάτερον ἄρα τῶν ΓΖ, ΩΨ στερεῶν πρὸς τὸ ΡΙ τὸν αὐτὸν ἔχει λόγον· ἴσον ἄρα ἐστὶ τὸ ΓΖ στερεὸν τῷ ΩΨ στερεῷ. ἀλλὰ τὸ ΩΨ τῷ ΑΕ ἐδείχθη ἴσον· καὶ τὸ ΑΕ ἄρα τῷ ΓΖ ἐστιν ἴσον. μὴ ἔστωσαν δὴ αἱ ἐφεστηκυῖαι αἱ ΑΗ, ΘΚ, ΒΕ, ΛΜ, ΓΝ, ΟΠ, ΔΖ, ΡΣ πρὸς ὀρθὰς ταῖς ΑΒ, ΓΔ βάσεσιν· λέγω πάλιν, ὅτι ἴσον τὸ ΑΕ στερεὸν τῷ ΓΖ στερεῷ. ἤχθωσαν γὰρ ἀπὸ τῶν Κ, Ε, Η, Μ, Π, Ζ, Ν, Σ σημείων ἐπὶ τὸ ὑποκείμενον ἐπίπεδον κάθετοι αἱ ΚΞ, ΕΤ, ΗΥ, ΜΦ, ΠΧ, ΖΨ, ΝΩ, ΣΙ, καὶ συμβαλλέτωσαν τῷ ἐπιπέδῳ κατὰ τὰ Ξ, Τ, Υ, Φ, Χ, Ψ, Ω, Ι σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΞΤ, ΞΥ, ΥΦ, ΤΦ, ΧΨ, ΧΩ, ΩΙ, ΙΨ. ἴσον δή ἐστι τὸ ΚΦ στερεὸν τῷ ΠΙ στερεῷ· ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΚΜ, ΠΣ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι πρὸς ὀρθάς εἰσι ταῖς βάσεσιν. ἀλλὰ τὸ μὲν ΚΦ στερεὸν τῷ ΑΕ στερεῷ ἐστιν ἴσον, τὸ δὲ ΠΙ τῷ ΓΖ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὔκ εἰσιν ἐπὶ τῶν αὐτῶν εὐθειῶν. καὶ τὸ ΑΕ ἄρα στερεὸν τῷ ΓΖ στερεῷ ἐστιν ἴσον. τὰ ἄρα ἐπὶ ἴσων βάσεων ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
32. Parallelepipedal solids which are of the same height are to one another as their bases. Let AB, CD be parallelepipedal solids of the same height; I say that the parallelepipedal solids AB, CD are to one another as their bases, that is, that, as the base AE is to the base CF, so is the solid AB to the solid CD. For let FH equal to AE be applied to FG, [] and on FH as base, and with the same height as that of CD, let the parallelepipedal solid GK be completed. Then the solid AB is equal to the solid GK; for they are on equal bases AE, FH and of the same height. [] And, since the parallelepipedal solid CK is cut by the plane DG which is parallel to opposite planes, therefore, as the base CF is to the base FH, so is the solid CD to the solid DH. [] But the base FH is equal to the base AE, and the solid GK to the solid AB; therefore also, as the base AE is to the base CF, so is the solid AB to the solid CD. Therefore etc. Q. E. D.
32. τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις. ἔστω ὑπὸ τὸ αὐτὸ ὕψος στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ· λέγω, ὅτι τὰ ΑΒ, ΓΔ στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις, τουτέστιν ὅτι ἐστὶν ὡς ἡ ΑΕ βάσις πρὸς τὴν ΓΖ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΔ στερεόν. παραβεβλήσθω γὰρ παρὰ τὴν ΖΗ τῷ ΑΕ ἴσον τὸ ΖΘ, καὶ ἀπὸ βάσεως μὲν τῆς ΖΘ, ὕψους δὲ τοῦ αὐτοῦ τῷ ΓΔ στερεὸν παραλληλεπίπεδον συμπεπληρώσθω τὸ ΗΚ. ἴσον δή ἐστι τὸ ΑΒ στερεὸν τῷ ΗΚ στερεῷ· ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΑΕ, ΖΘ καὶ ὑπὸ τὸ αὐτὸ ὕψος. καὶ ἐπεὶ στερεὸν παραλληλεπίπεδον τὸ ΓΚ ἐπιπέδῳ τῷ ΔΗ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ἄρα ὡς ἡ ΓΖ βάσις πρὸς τὴν ΖΘ βάσιν, οὕτως τὸ ΓΔ στερεὸν πρὸς τὸ ΔΘ στερεόν. ἴση δὲ ἡ μὲν ΖΘ βάσις τῇ ΑΕ βάσει, τὸ δὲ ΗΚ στερεὸν τῷ ΑΒ στερεῷ· ἔστιν ἄρα καὶ ὡς ἡ ΑΕ βάσις πρὸς τὴν ΓΖ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΔ στερεόν. τὰ ἄρα ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις· ὅπερ ἔδει δεῖξαι.
33. Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides. Let AB, CD be similar parallelepipedal solids, and let AE be the side corresponding to CF; I say that the solid AB has to the solid CD the ratio triplicate of that which AE has to CF. For let EK, EL, EM be produced in a straight line with AE, GE, HE, let EK be made equal to CF, EL equal to FN, and further EM equal to FR, and let the parallelogram KL and the solid KP be completed. Now, since the two sides KE, EL are equal to the two sides CF, FN, while the angle KEL is also equal to the angle CFN, inasmuch as the angle AEG is also equal to the angle CFN because of the similarity of the solids AB, CD, therefore the parallelogram KL is equal ltand similargt to the parallelogram CN. For the same reason the parallelogram KM is also equal and similar to CR, and further EP to DF; therefore three parallelograms of the solid KP are equal and similar to three parallelograms of the solid CD. But the former three parallelograms are equal and similar to their opposites, and the latter three to their opposites; [] therefore the whole solid KP is equal and similar to the whole solid CD. [] Let the parallelogram GK be completed, and on the parallelograms GK, KL as bases, and with the same height as that of AB, let the solids EO, LQ be completed. Then since; owing to the similarity of the solids AB, CD, as AE is to CF, so is EG to FN, and EH to FR, while CF is equal to EK, FN to EL, and FR to EM, therefore, as AE is to EK, so is GE to EL, and HE to EM. But, as AE is to EK, so is AG to the parallelogram GK, as GE is to EL, so is GK to KL, and, as HE is to EM, so is QE to KM; [] therefore also, as the parallelogram AG is to GK, so is GK to KL, and QE to KM. But, as AG is to GK, so is the solid AB to the solid EO, as GK is to KL, so is the solid OE to the solid QL, and, as QE is to KM, so is the solid QL to the solid KP; [] therefore also, as the solid AB is to EO, so is EO to QL, and QL to KP. But, if four magnitudes be continuously proportional, the first has to the fourth the ratio triplicate of that which it has to the second; [] therefore the solid AB has to KP the ratio triplicate of that which AB has to EO. But, as AB is to EO, so is the parallelogram AG to GK, and the straight line AE to EK []; hence the solid AB has also to KP the ratio triplicate of that which AE has to EK. But the solid KP is equal to the solid CD, and the straight line EK to CF; therefore the solid AB has also to the solid CD the ratio triplicate of that which the corresponding side of it, AE, has to the corresponding side CF. Therefore etc. Q. E. D. Porism. From this it is manifest that, if four straight lines be ltcontinuouslygt proportional, as the first is to the fourth, so will a parallelepipedal solid on the first be to the similar and similarly described parallelepipedal solid on the second, inasmuch as the first has to the fourth the ratio triplicate of that which it has to the second.
33. τὰ ὅμοια στερεὰ παραλληλεπίπεδα πρὸς ἄλληλα ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ἔστω ὅμοια στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ, ὁμόλογος δὲ ἔστω ἡ ΑΕ τῇ ΓΖ· λέγω, ὅτι τὸ ΑΒ στερεὸν πρὸς τὸ ΓΔ στερεὸν τριπλασίονα λόγον ἔχει, ἤπερ ἡ ΑΕ πρὸς τὴν ΓΖ. ἐκβεβλήσθωσαν γὰρ ἐπʼ εὐθείας ταῖς ΑΕ, ΗΕ, ΘΕ αἱ ΕΚ, ΕΛ, ΕΜ, καὶ κείσθω τῇ μὲν ΓΖ ἴση ἡ ΕΚ, τῇ δὲ ΖΝ ἴση ἡ ΕΛ, καὶ ἔτι τῇ ΖΡ ἴση ἡ ΕΜ, καὶ συμπεπληρώσθω τὸ ΚΛ παραλληλόγραμμον καὶ τὸ ΚΟ στερεόν. καὶ ἐπεὶ δύο αἱ ΚΕ, ΕΛ δυσὶ ταῖς ΓΖ, ΖΝ ἴσαι εἰσίν, ἀλλὰ καὶ γωνία ἡ ὑπὸ ΚΕΛ γωνίᾳ τῇ ὑπὸ ΓΖΝ ἐστιν ἴση, ἐπειδήπερ καὶ ἡ ὑπὸ ΑΕΗ τῇ ὑπὸ ΓΖΝ ἐστιν ἴση διὰ τὴν ὁμοιότητα τῶν ΑΒ, ΓΔ στερεῶν, ἴσον ἄρα ἐστὶ [καὶ ὅμοιον] τὸ ΚΛ παραλληλόγραμμον τῷ ΓΝ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν ΚΜ παραλληλόγραμμον ἴσον ἐστὶ καὶ ὅμοιον τῷ ΓΡ [παραλληλογράμμῳ] καὶ ἔτι τὸ ΕΟ τῷ ΔΖ· τρία ἄρα παραλληλόγραμμα τοῦ ΚΟ στερεοῦ τρισὶ παραλληλογράμμοις τοῦ ΓΔ στερεοῦ ἴσα ἐστὶ καὶ ὅμοια. ἀλλὰ τὰ μὲν τρία τρισὶ τοῖς ἀπεναντίον ἴσα ἐστὶ καὶ ὅμοια, τὰ δὲ τρία τρισὶ τοῖς ἀπεναντίον ἴσα ἐστὶ καὶ ὅμοια· ὅλον ἄρα τὸ ΚΟ στερεὸν ὅλῳ τῷ ΓΔ στερεῷ ἴσον ἐστὶ καὶ ὅμοιον. συμπεπληρώσθω τὸ ΗΚ παραλληλόγραμμον, καὶ ἀπὸ βάσεων μὲν τῶν ΗΚ, ΚΛ παραλληλογράμμων, ὕψους δὲ τοῦ αὐτοῦ τῷ ΑΒ στερεὰ συμπεπληρώσθω τὰ ΕΞ, ΛΠ. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν ΑΒ, ΓΔ στερεῶν ἐστιν ὡς ἡ ΑΕ πρὸς τὴν ΓΖ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΝ, καὶ ἡ ΕΘ πρὸς τὴν ΖΡ, ἴση δὲ ἡ μέν ΓΖ τῇ ΕΚ, ἡ δὲ ΖΝ τῇ ΕΛ, ἡ δὲ ΖΡ τῇ ΕΜ, ἔστιν ἄρα ὡς ἡ ΑΕ πρὸς τὴν ΕΚ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΛ καὶ ἡ ΘΕ πρὸς τὴν ΕΜ. ἀλλʼ ὡς μὲν ἡ ΑΕ πρὸς τὴν ΕΚ, οὕτως τὸ ΑΗ [παραλληλόγραμμον] πρὸς τὸ ΗΚ παραλληλόγραμμον, ὡς δὲ ἡ ΗΕ πρὸς τὴν ΕΛ, οὕτως τὸ ΗΚ πρὸς τὸ ΚΛ, ὡς δὲ ἡ ΘΕ πρὸς ΕΜ, οὕτως τὸ ΠΕ πρὸς τὸ ΚΜ· καὶ ὡς ἄρα τὸ ΑΗ παραλληλόγραμμον πρὸς τὸ ΗΚ, οὕτως τὸ ΗΚ πρὸς τὸ ΚΛ καὶ τὸ ΠΕ πρὸς τὸ ΚΜ. ἀλλʼ ὡς μὲν τὸ ΑΗ πρὸς τὸ ΗΚ, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΕΞ στερεόν, ὡς δὲ τὸ ΗΚ πρὸς τὸ ΚΛ, οὕτως τὸ ΞΕ στερεὸν πρὸς τὸ ΠΛ στερεόν, ὡς δὲ τὸ ΠΕ πρὸς τὸ ΚΜ, οὕτως τὸ ΠΛ στερεὸν πρὸς τὸ ΚΟ στερεόν· καὶ ὡς ἄρα τὸ ΑΒ στερεὸν πρὸς τὸ ΕΞ, οὕτως τὸ ΕΞ πρὸς τὸ ΠΛ καὶ τὸ ΠΛ πρὸς τὸ ΚΟ. ἐὰν δὲ τέσσαρα μεγέθη κατὰ τὸ συνεχὲς ἀνάλογον ᾖ, τὸ πρῶτον πρὸς τὸ τέταρτον τριπλασίονα λόγον ἔχει ἤπερ πρὸς τὸ δεύτερον· τὸ ΑΒ ἄρα στερεὸν πρὸς τὸ ΚΟ τριπλασίονα λόγον ἔχει ἤπερ τὸ ΑΒ πρὸς τὸ ΕΞ. ἀλλʼ ὡς τὸ ΑΒ πρὸς τὸ ΕΞ, οὕτως τὸ ΑΗ παραλληλόγραμμον πρὸς τὸ ΗΚ καὶ ἡ ΑΕ εὐθεῖα πρὸς τὴν ΕΚ· ὥστε καὶ τὸ ΑΒ στερεὸν πρὸς τὸ ΚΟ τριπλασίονα λόγον ἔχει ἤπερ ἡ ΑΕ πρὸς τὴν ΕΚ. ἴσον δὲ τὸ [μὲν] ΚΟ στερεὸν τῷ ΓΔ στερεῷ, ἡ δὲ ΕΚ εὐθεῖα τῇ ΓΖ· καὶ τὸ ΑΒ ἄρα στερεὸν πρὸς τὸ ΓΔ στερεὸν τριπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος αὐτοῦ πλευρὰ ἡ ΑΕ πρὸς τὴν ὁμόλογον πλευρὰν τὴν ΓΖ. τὰ ἄρα ὅμοια στερεὰ παραλληλεπίπεδα ἐν τριπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν· ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, ἔσται ὡς ἡ πρώτη πρὸς τὴν τετάρτην, οὕτω τὸ ἀπὸ τῆς πρώτης στερεὸν παραλληλεπίπεδον πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον, ἐπείπερ καὶ ἡ πρώτη πρὸς τὴν τετάρτην τριπλασίονα λόγον ἔχει ἤπερ πρὸς τὴν δευτέραν.
34. In equal parallelepipedal solids the bases are reciprocally proportional to the heights; and those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal.
34. τῶν ἴσων στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν· καὶ ὧν στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἴσα ἐστὶν ἐκεῖνα. ἔστω ἴσα στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ·
Let AB, CD be equal parallelepipedal solids; I say that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB.
λέγω, ὅτι τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων 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. ἔστωσαν γὰρ πρότερον αἱ ἐφεστηκυῖαι αἱ ΑΗ, ΕΖ, ΛΒ, ΘΚ, ΓΜ, ΝΞ, ΟΔ, ΠΡ πρὸς ὀρθὰς ταῖς βάσεσιν αὐτῶν·
First, let the sides which stand up, namely AG, EF, LB, HK, CM, NO, PD, QR, be at right angles to their bases; I say that, as the base EH is to the base NQ, so is CM to AG.
λέγω, ὅτι ἐστὶν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΓΜ πρὸς τὴν ΑΗ. εἰ μὲν οὖν ἴση ἐστιν ἡ ΕΘ βάσις τῇ ΝΠ βάσει, ἔστι δὲ καὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ ἴσον, ἔσται καὶ ἡ ΓΜ τῇ ΑΗ ἴση.
If now the base EH is equal to the base NQ, while the solid AB is also equal to the solid CD, CM will also be equal to AG.
τὰ γὰρ ὑπὸ τὸ αὐτὸ ὕψος στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις [εἰ γὰρ τῶν ΕΘ, ΝΠ βάσεων ἴσων οὐσῶν μὴ εἴη τὰ ΑΗ, ΓΜ ὕψη ἴσα, οὐδʼ ἄρα τὸ ΑΒ στερεὸν ἴσον ἔσται τῷ ΓΔ.
For parallelepipedal solids of the same height are to one another as the bases; [] and, as the base EH is to NQ, so will CM be to AG, and it is manifest that in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.
ὑπόκειται δὲ ἴσον· οὐκ ἄρα ἄνισόν ἐστι τὸ ΓΜ ὕψος τῷ ΑΗ ὕψει· ἴσον ἄρα]. καὶ ἔσται ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ, οὕτως ἡ ΓΜ πρὸς τὴν ΑΗ, καὶ φανερόν, ὅτι τῶν ΑΒ, ΓΔ στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν.
Next, let the base EH not be equal to the base NQ, but let EH be greater.
μὴ ἔστω δὴ ἴση ἡ ΕΘ βάσις τῇ ΝΠ βάσει, ἀλλʼ ἔστω μείζων ἡ ΕΘ.
Now the solid AB is equal to the solid CD; therefore CM is also greater than AG.
ἔστι δὲ καὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ ἴσον·
Let then CT be made equal to AG, and let the parallelepipedal solid VC be completed on NQ as base and with CT as height.
μείζων ἄρα ἐστὶ καὶ ἡ ΓΜ τῆς ΑΗ [εἰ γὰρ μή, οὐδʼ ἄρα πάλιν τὰ ΑΒ, ΓΔ στερεὰ ἴσα ἔσται· ὑπόκειται δὲ ἴσα].
Now, since the solid AB is equal to the solid CD, and CV is outside them, while equals have to the same the same ratio, [] therefore, as the solid AB is to the solid CV, so is the solid CD to the solid CV.
κείσθω οὖν τῇ ΑΗ ἴση ἡ ΓΤ, καὶ συμπεπληρώσθω ἀπὸ βάσεως μὲν τῆς ΝΠ, ὕψους δὲ τοῦ ΓΤ, στερεὸν παραλληλεπίπεδον τὸ ΦΓ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ, ἔξωθεν δὲ τὸ ΓΦ, τὰ δὲ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν.
But, as the solid AB is to the solid CV, so is the base EH to the base NQ, for the solids AB, CV are of equal height; [] and, as the solid CD is to the solid CV, so is the base MQ to the base TQ [] and CM to CT []; therefore also, as the base EH is to the base NQ, so is MC to CT.
ἀλλʼ ὡς μὲν τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν· ἰσοϋψῆ γὰρ τὰ ΑΒ, ΓΦ στερεά· ὡς δὲ τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν, οὕτως ἡ ΜΠ βάσις πρὸς τὴν ΤΠ βάσιν καὶ ἡ ΓΜ πρὸς τὴν ΓΤ·
But CT is equal to AG; therefore also, as the base EH is to the base NQ, so is MC to AG.
καὶ ὡς ἄρα ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΜΓ πρὸς τὴν ΓΤ. ἴση δὲ ἡ ΓΤ τῇ ΑΗ·
Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the heights.
καὶ ὡς ἄρα ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΜΓ πρὸς τὴν ΑΗ. τῶν ΑΒ, ΓΔ ἄρα στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν.
Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD. Let the sides which stand up be again at right angles to the bases.
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.
Now, if the base EH is equal to the base NQ, and, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB, therefore the height of the solid CD is also equal to the height of the solid AB.
ἔστωσαν [γὰρ] πάλιν αἱ ἐφεστηκυῖαι πρὸς ὀρθὰς ταῖς βάσεσιν, καὶ εἰ μὲν ἴση ἐστὶν ἡ ΕΘ βάσις τῇ ΝΠ βάσει, καί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΓΔ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος, ἴσον ἄρα ἐστὶ καὶ τὸ τοῦ ΓΔ στερεοῦ ὕψος τῷ τοῦ ΑΒ στερεοῦ ὕψει. τὰ δὲ ἐπὶ ἴσων βάσεων στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος ἴσα ἀλλήλοις ἐστίν·
But parallelepipedal solids on equal bases and of the same height are equal to one another; [] therefore the solid AB is equal to the solid CD.
ἴσον ἄρα ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ. μὴ ἔστω δὴ ἡ ΕΘ βάσις τῇ ΝΠ [βάσει] ἴση, ἀλλʼ ἔστω μείζων ἡ ΕΘ·
Next, let the base EH not be equal to the base NQ, but let EH be greater; therefore the height of the solid CD is also greater than the height of the solid AB, that is, CM is greater than AG.
μεῖζον ἄρα ἐστὶ καὶ τὸ τοῦ ΓΔ στερεοῦ ὕψος τοῦ τοῦ ΑΒ στερεοῦ ὕψους, τουτέστιν ἡ ΓΜ τῆς ΑΗ. κείσθω τῇ ΑΗ ἴση πάλιν ἡ ΓΤ, καὶ συμπεπληρώσθω ὁμοίως τὸ ΓΦ στερεόν.
Let CT be again made equal to AG, and let the solid CV be similarly completed.
ἐπεί ἐστιν ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΜΓ πρὸς τὴν ΑΗ, ἴση δὲ ἡ ΑΗ τῇ ΓΤ, ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως ἡ ΓΜ πρὸς τὴν ΓΤ.
Since, as the base EH is to the base NQ, so is MC to AG, while AG is equal to CT, therefore, as the base EH is to the base NQ, so is CM to CT.
ἀλλʼ ὡς μὲν ἡ ΕΘ [βάσις] πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ ΑΒ στερεὸν πρὸς τὸ ΓΦ στερεόν· ἰσοϋψῆ γάρ ἐστι τὰ ΑΒ, ΓΦ στερεά·
But, as the base EH is to the base NQ, so is the solid AB to the solid CV, for the solids AB, CV are of equal height; [] and, as CM is to CT, so is the base MQ to the base QT [] and the solid CD to the solid CV. []
ὡς δὲ ἡ ΓΜ πρὸς τὴν ΓΤ, οὕτως ἥ τε ΜΠ βάσις πρὸς τὴν ΠΤ βάσιν καὶ τὸ ΓΔ στερεὸν πρὸς τὸ ΓΦ στερεόν. 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.
Therefore also, as the solid AB is to the solid CV, so is the solid CD to the solid CV; therefore each of the solids AB, CD has to CV the same ratio.
ἴσον ἄρα ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ [ὅπερ ἔδει δεῖξαι].
Therefore the solid AB is equal to the solid CD. []
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·
Now let the sides which stand up, FE, BL, GA, HK, ON, DP, MC, RQ, not be at right angles to their bases; let perpendiculars be drawn from the points F, G, B, K, O, M, D, R to the planes through EH, NQ, and let them meet the planes at S, T, U, V, W, X, Y, a, and let the solids FV, Oa be completed; I say that, in this case too, if the solids AB, CD are equal, the bases are reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB.
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. 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 [ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν]· τὸ δὲ ΓΔ στερεὸν τῷ ΔΨ ἐστιν ἴσον·
Since the solid AB is equal to the solid CD, while AB is equal to BT, for they are on the same base FK and of the same height; [] and the solid CD is equal to DX, for they are again on the same base RO and of the same height; [id.] therefore the solid BT is also equal to the solid DX.
ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς ΡΞ καὶ ὑπὸ τὸ αὐτὸ ὕψος [ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν]· καὶ τὸ ΒΤ ἄρα στερεὸν τῷ ΔΨ στερεῷ ἴσον ἐστίν [τῶν δὲ ἴσων στερεῶν παραλληλεπιπέδων, ὧν τὰ ὕψη πρὸς ὀρθάς ἐστι ταῖς βάσεσιν αὐτῶν,] [ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν].
Therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT. [Part 1.]
ἔστιν ἄρα ὡς ἡ ΖΚ βάσις πρὸς τὴν ΞΡ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ στερεοῦ ὕψος. ἴση δὲ ἡ μὲν ΖΚ βάσις τῇ ΕΘ βάσει, ἡ δὲ ΞΡ βάσις τῇ ΝΠ βάσει·
But the base FK is equal to the base EH, and the base OR to the base NQ; therefore, as the base EH is to the base NQ, so is the height of the solid DX to the height of the solid BT.
ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΔΨ στερεοῦ ὕψος πρὸς τὸ τοῦ ΒΤ στερεοῦ ὕψος. τὰ δʼ αὐτὰ ὕψη ἐστὶ τῶν ΔΨ, ΒΤ στερεῶν καὶ τῶν ΔΓ, ΒΑ·
But the solids DX, BT and the solids DC, BA have the same heights respectively; therefore, as the base EH is to the base NQ, so is the height of the solid DC to the height of the solid AB.
ἔστιν ἄρα ὡς ἡ ΕΘ βάσις πρὸς τὴν ΝΠ βάσιν, οὕτως τὸ τοῦ ΔΓ στερεοῦ ὕψος πρὸς τὸ τοῦ ΑΒ στερεοῦ ὕψος. τῶν ΑΒ, ΓΔ ἄρα στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν.
Therefore in the parallelepipedal solids AB, CD the bases are reciprocally proportional to the 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·
Again, in the parallelepipedal solids AB, CD let the bases be reciprocally proportional to the heights, that is, as the base EH is to the base NQ, so let the height of the solid CD be to the height of the solid AB; I say that the solid AB is equal to the solid CD.
λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ στερεὸν τῷ ΓΔ στερεῷ. 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 construction, since, as the base EH is to the base NQ, so is the height of the solid CD to the height of the solid AB, while the base EH is equal to the base FK, and NQ to OR, therefore, as the base FK is to the base OR, so is the height of the solid CD 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 solids AB, CD and BT, DX have the same heights respectively; therefore, as the base FK is to the base OR, so is the height of the solid DX to the height of the solid BT.
τῶν ΒΤ, ΔΨ ἄρα στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν [ὧν δὲ στερεῶν παραλληλεπιπέδων τὰ ὕψη πρὸς ὀρθάς ἐστι ταῖς βάσεσιν αὐτῶν, ἀντιπεπόνθασι δὲ αἱ βάσεις τοῖς ὕψεσιν, ἴσα ἐστὶν ἐκεῖνα]· ἴσον ἄρα ἐστὶ τὸ ΒΤ στερεὸν τῷ ΔΨ στερεῷ.
Therefore in the parallelepipedal solids BT, DX the bases are reciprocally proportional to the heights; therefore the solid BT is equal to the solid DX. [Part 1.]
ἀλλὰ τὸ μὲν ΒΤ τῷ ΒΑ ἴσον ἐστίν· ἐπί τε γὰρ τῆς αὐτῆς βάσεως [εἰσι] τῆς ΖΚ καὶ ὑπὸ τὸ αὐτὸ ὕψος [ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν].
But BT is equal to BA, for they are on the same base FK and of the same height; [] and the solid DX is equal to the solid DC. [id.]
τὸ δὲ ΔΨ στερεὸν τῷ ΔΓ στερεῷ ἴσον ἐστίν [ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς ΞΡ καὶ ὑπὸ τὸ αὐτὸ ὕψος καὶ οὐκ ἐν ταῖς αὐταῖς εὐθείαις].
Therefore the solid AB is also equal to the solid CD. Q. E. D.
καὶ τὸ ΑΒ ἄρα στερεὸν τῷ ΓΔ στερεῷ ἐστιν ἴσον· ὅπερ ἔδει δεῖξαι.
35. If there be two equal plane angles, and on their vertices there be set up elevated straight lines containing equal angles with the original straight lines respectively, if on the elevated straight lines points be taken at random and perpendiculars be drawn from them to the planes in which the original angles are, and if from the points so arising in the planes straight lines be joined to the vertices of the original angles, they will contain, with the elevated straight lines, equal angles. Let the angles BAC, EDF be two equal rectilineal angles, and from the points A, D let the elevated straight lines AG, DM be set up containing, with the original straight lines, equal angles respectively, namely, the angle MDE to the angle GAB and the angle MDF to the angle GAC, let points G, M be taken at random on AG, DM, let GL, MN be drawn from the points G, M perpendicular to the planes through BA, AC and ED, DF, and let them meet the planes at L, N, and let LA, ND be joined; I say that the angle GAL is equal to the angle MDN. Let AH be made equal to DM, and let HK be drawn through the point H parallel to GL. But GL is perpendicular to the plane through BA, AC; therefore HK is also perpendicular to the plane through. BA, AC. [] From the points K, N let KC, NF, KB, NE be drawn perpendicular to the straight lines AC, DF, AB, DE, and let HC, CB, MF, FE be joined. Since the square on HA is equal to the squares on HK, KA, and the squares on KC, CA are equal to the square on KA, [] therefore the square on HA is also equal to the squares on HK, KC, CA. But the square on HC is equal to the squares on HK, KC; [] therefore the square on HA is equal to the squares on HC, CA. Therefore the angle HCA is right. [] For the same reason the angle DFM is also right. Therefore the angle ACH is equal to the angle DFM. But the angle HAC is also equal to the angle MDF. Therefore MDF, HAC are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that subtending one of the equal angles, that is, HA equal to MD; therefore they will also have the remaining sides equal to the remaining sides respectively. [] Therefore AC is equal to DF. Similarly we can prove that AB is also equal to DE. Since then AC is equal to DF, and AB to DE, the two sides CA, AB are equal to the two sides FD, DE. But the angle CAB is also equal to the angle FDE; therefore the base BC is equal to the base EF, the triangle to the triangle, and the remaining angles to the remaining angles; [] therefore the angle ACB is equal to the angle DFE. But the right angle ACK is also equal to the right angle DFN; therefore the remaining angle BCK is also equal to the remaining angle EFN. For the same reason the angle CBK is also equal to the angle FEN. Therefore BCK, EFN are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely, that adjacent to the equal angles, that is, BC equal to EF; therefore they will also have the remaining sides equal to the remaining sides. [] Therefore CK is equal to FN. But AC is also equal to DF; therefore the two sides AC, CK are equal to the two sides DF, FN; and they contain right angles. Therefore the base AK is equal to the base DN. [] And, since AH is equal to DM, the square on AH is also equal to the square on DM. But the squares on AK, KH are equal to the square on AH, for the angle AKH is right; [] and the squares on DN, NM are equal to the square on DM, for the angle DNM is right; [] therefore the squares on AK, KH are equal to the squares on DN, NM; and of these the square on AK is equal to the square on DN; therefore the remaining square on KH is equal to the square on NM; therefore HK is equal to MN. And, since the two sides HA, AK are equal to the two sides MD, DN respectively, and the base HK was proved equal to the base MN, therefore the angle HAK is equal to the angle MDN. [] Therefore etc. Porism. From this it is manifest that, if there be two equal plane angles, and if there be set up on them elevated straight lines which are equal and contain equal angles with the original straight lines respectively, the perpendiculars drawn from their extremities to the planes in which are the original angles are equal to one another. Q. E. D.
35. ἐὰν ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι, ἐπὶ δὲ τῶν κορυφῶν αὐτῶν μετέωροι εὐθεῖαι ἐπισταθῶσιν ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, ἐπὶ δὲ τῶν μετεώρων ληφθῇ τυχόντα σημεῖα, καὶ ἀπʼ αὐτῶν ἐπὶ τὰ ἐπίπεδα, ἐν οἷς εἰσιν αἱ ἐξ ἀρχῆς γωνίαι, κάθετοι ἀχθῶσιν, ἀπὸ δὲ τῶν γενομένων σημείων ἐν τοῖς ἐπιπέδοις ἐπὶ τὰς ἐξ ἀρχῆς γωνίας ἐπιζευχθῶσιν εὐθεῖαι, ἴσας γωνίας περιέξουσι μετὰ τῶν μετεώρων. ἔστωσαν δύο γωνίαι εὐθύγραμμοι ἴσαι αἱ ὑπὸ ΒΑΓ, ΕΔΖ, ἀπὸ δὲ τῶν Α, Δ σημείων μετέωροι εὐθεῖαι ἐφεστάτωσαν αἱ ΑΗ, ΔΜ ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, τὴν μὲν ὑπὸ ΜΔΕ τῇ ὑπὸ ΗΑΒ, τὴν δὲ ὑπὸ ΜΔΖ τῇ ὑπὸ ΗΑΓ, καὶ εἰλήφθω ἐπὶ τῶν ΑΗ, ΔΜ τυχόντα σημεῖα τὰ Η, Μ, καὶ ἤχθωσαν ἀπὸ τῶν Η, Μ σημείων ἐπὶ τὰ διὰ τῶν ΒΑΓ, ΕΔΖ ἐπίπεδα κάθετοι αἱ ΗΛ, ΜΝ, καὶ συμβαλλέτωσαν τοῖς ἐπιπέδοις κατὰ τὰ Ν, Λ, καὶ ἐπεζεύχθωσαν αἱ ΛΑ, ΝΔ· λέγω, ὅτι ἴση ἐστὶν ἡ ὑπὸ ΗΑΛ γωνία τῇ ὑπὸ ΜΔΝ γωνίᾳ. κείσθω τῇ ΔΜ ἴση ἡ ΑΘ, καὶ ἤχθω διὰ τοῦ Θ σημείου τῇ ΗΛ παράλληλος ἡ ΘΚ. ἡ δὲ ΗΛ κάθετός ἐστιν ἐπὶ τὸ διὰ τῶν ΒΑΓ ἐπίπεδον· καὶ ἡ ΘΚ ἄρα κάθετός ἐστιν ἐπὶ τὸ διὰ τῶν ΒΑΓ ἐπίπεδον. ἤχθωσαν ἀπὸ τῶν Κ, Ν σημείων ἐπὶ τὰς ΑΒ, ΑΓ, ΔΖ, ΔΕ εὐθείας κάθετοι αἱ ΚΓ, ΝΖ, ΚΒ, ΝΕ, καὶ ἐπεζεύχθωσαν αἱ ΘΓ, ΓΒ, ΜΖ, ΖΕ. ἐπεὶ τὸ ἀπὸ τῆς ΘΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΑ, τῷ δὲ ἀπὸ τῆς ΚΑ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΚΓ, ΓΑ, καὶ τὸ ἀπὸ τῆς ΘΑ ἄρα ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΚ, ΚΓ, ΓΑ. τοῖς δὲ ἀπὸ τῶν ΘΚ, ΚΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΘΓ· τὸ ἄρα ἀπὸ τῆς ΘΑ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΘΓ, ΓΑ. ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΘΓΑ γωνία. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΔΖΜ γωνία ὀρθή ἐστιν. ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΓΘ γωνία τῇ ὑπὸ ΔΖΜ. ἔστι δὲ καὶ ἡ ὑπὸ ΘΑΓ τῇ ὑπὸ ΜΔΖ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΜΔΖ, ΘΑΓ δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν τὴν ΘΑ τῇ ΜΔ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ. ἴση ἄρα ἐστὶν ἡ ΑΓ τῇ ΔΖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΑΒ τῇ ΔΕ ἐστιν ἴση [οὕτως· ἐπεζεύχθωσαν αἱ ΘΒ, ΜΕ. καὶ ἐπεὶ τὸ ἀπὸ τῆς ΑΘ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΚ, ΚΘ, τῷ δὲ ἀπὸ τῆς ΑΚ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΚ, τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΚ, ΚΘ ἴσα ἐστὶ τῷ ἀπὸ ΑΘ. ἀλλὰ τοῖς ἀπὸ τῶν ΒΚ, ΚΘ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΘ· ὀρθὴ γὰρ ἡ ὑπὸ ΘΚΒ γωνία διὰ τὸ καὶ τὴν ΘΚ κάθετον εἶναι ἐπὶ τὸ ὑποκείμενον ἐπίπεδον· τὸ ἄρα ἀπὸ τῆς ΑΘ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΑΒ, ΒΘ· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΑΒΘ γωνία. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΔΕΜ γωνία ὀρθή ἐστιν. ἔστι δὲ καὶ ἡ ὑπὸ ΒΑΘ γωνία τῇ ὑπὸ ΕΔΜ ἴση· ὑπόκεινται γάρ· καὶ ἔστιν ἡ ΑΘ τῇ ΔΜ ἴση· ἴση ἄρα ἐστὶ καὶ ἡ ΑΒ τῇ ΔΕ ]. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΓ τῇ ΔΖ, ἡ δὲ ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΓΑ, ΑΒ δυσὶ ταῖς ΖΔ, ΔΕ ἴσαι εἰσίν. ἀλλὰ καὶ γωνία ἡ ὑπὸ ΓΑΒ γωνίᾳ τῇ ὑπὸ ΖΔΕ ἐστιν ἴση· βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστὶ καὶ τὸ τρίγωνον τῷ τριγώνῳ καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις· ἴση ἄρα ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἔστι δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΓΚ ὀρθῇ τῇ ὑπὸ ΔΖΝ ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΓΚ λοιπῇ τῇ ὑπὸ ΕΖΝ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΒΚ τῇ ὑπὸ ΖΕΝ ἐστιν ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΒΓΚ, ΕΖΝ [τὰς] δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΒΓ τῇ ΕΖ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν. ἴση ἄρα ἐστὶν ἡ ΓΚ τῇ ΖΝ. ἔστι δὲ καὶ ἡ ΑΓ τῇ ΔΖ ἴση· δύο δὴ αἱ ΑΓ, ΓΚ δυσὶ ταῖς ΔΖ, ΖΝ ἴσαι εἰσίν· καὶ ὀρθὰς γωνίας περιέχουσιν. βάσις ἄρα ἡ ΑΚ βάσει τῇ ΔΝ ἴση ἐστίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΘ τῇ ΔΜ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΘ τῷ ἀπὸ τῆς ΔΜ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΘ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΚ, ΚΘ· ὀρθὴ γὰρ ἡ ὑπὸ ΑΚΘ· τῷ δὲ ἀπὸ τῆς ΔΜ ἴσα τὰ ἀπὸ τῶν ΔΝ, ΝΜ· ὀρθὴ γὰρ ἡ ὑπὸ ΔΝΜ· τὰ ἄρα ἀπὸ τῶν ΑΚ, ΚΘ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΔΝ, ΝΜ, ὧν τὸ ἀπὸ τῆς ΑΚ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΝ· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΚΘ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΝΜ· ἴση ἄρα ἡ ΘΚ τῇ ΜΝ. καὶ ἐπεὶ δύο αἱ ΘΑ, ΑΚ δυσὶ ταῖς ΜΔ, ΔΝ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ βάσις ἡ ΘΚ βάσει τῇ ΜΝ ἐδείχθη ἴση, γωνία ἄρα ἡ ὑπὸ ΘΑΚ γωνίᾳ τῇ ὑπὸ ΜΔΝ ἐστιν ἴση. ἐὰν ἄρα ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι καὶ τὰ ἑξῆς τῆς προτάσεως [ὅπερ ἔδει δεῖξαι]. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι, ἐπισταθῶσι δὲ ἐπʼ αὐτῶν μετέωροι εὐθεῖαι ἴσαι ἴσας γωνίας περιέχουσαι μετὰ τῶν ἐξ ἀρχῆς εὐθειῶν ἑκατέραν ἑκατέρᾳ, αἱ ἀπʼ αὐτῶν κάθετοι ἀγόμεναι ἐπὶ τὰ ἐπίπεδα, ἐν οἷς εἰσιν αἱ ἐξ ἀρχῆς γωνίαι, ἴσαι ἀλλήλαις εἰσίν. ὅπερ ἔδει δεῖξαι.