§11.prop.24ἐὰν στερεὸν ὑπὸ παραλλήλων ἐπιπέδων περιέχηται, τὰ ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά ἐστιν.
If a solid is contained by parallel planes, its opposite planes are equal and parallelograms.
στερεὸν γὰρ τὸ ΓΔΘΗ ὑπὸ παραλλήλων ἐπιπέδων περιεχέσθω τῶν ΑΓ, ΗΖ, ΑΘ, ΔΖ, ΒΖ, ΑΕ· λέγω, ὅτι τὰ ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά ἐστιν.
For let the solid CDHG be contained by parallel planes AC, GF, AH, DF, BF, AE; I say that its opposite planes are equal and parallelograms.
ἐπεὶ γὰρ δύο ἐπίπεδα παράλληλα τὰ ΒΗ, ΓΕ ὑπὸ ἐπιπέδου τοῦ ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν.
For since two parallel planes BG, CE are cut by the plane AC, their common sections are parallel.
παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΔΓ. πάλιν, ἐπεὶ δύο ἐπίπεδα παράλληλα τὰ ΒΖ, ΑΕ ὑπὸ ἐπιπέδου τοῦ ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν.
Therefore AB is parallel to DC. Again, since two parallel planes BF, AE are cut by the plane AC, their common sections are parallel.
παράλληλος ἄρα ἐστὶν ἡ ΒΓ τῇ ΑΔ. ἐδείχθη δὲ καὶ ἡ ΑΒ τῇ ΔΓ παράλληλος· παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΓ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἕκαστον τῶν ΔΖ, ΖΗ, ΗΒ, ΒΖ, ΑΕ παραλληλόγραμμόν ἐστιν.
Therefore BC is parallel to AD. But AB was also shown parallel to DC; therefore AC is a parallelogram. Similarly indeed we shall show that each of DF, FG, GB, BF, AE is also a parallelogram.
ἐπεζεύχθωσαν αἱ ΑΘ, ΔΖ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ μὲν ΑΒ τῇ ΔΓ, ἡ δὲ ΒΘ τῇ ΓΖ, δύο δὴ αἱ ΑΒ, ΒΘ ἁπτόμεναι ἀλλήλων παρὰ δύο εὐθείας τὰς ΔΓ, ΓΖ ἁπτομένας ἀλλήλων εἰσὶν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ· ἴσας ἄρα γωνίας περιέξουσιν· ἴση ἄρα ἡ ὑπὸ ΑΒΘ γωνία τῇ ὑπὸ ΔΓΖ. καὶ ἐπεὶ δύο αἱ ΑΒ, ΒΘ δυσὶ ταῖς ΔΓ, ΓΖ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ ΑΒΘ γωνίᾳ τῇ ὑπὸ ΔΓΖ ἐστιν ἴση, βάσις ἄρα ἡ ΑΘ βάσει τῇ ΔΖ ἐστιν ἴση, καὶ τὸ ΑΒΘ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἴσον ἐστίν.
Let AH, DF be joined. And since AB is parallel to DC, and BH to CF, the two straight lines AB, BH meeting one another are parallel to two straight lines DC, CF meeting one another, which are not in the same plane; therefore they will contain equal angles; therefore the angle ABH is equal to the angle DCF. And since the two straight lines AB, BH are equal to the two straight lines DC, CF, and the angle ABH is equal to the angle DCF, therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DCF.
καί ἐστι τοῦ μὲν ΑΒΘ διπλάσιον τὸ ΒΗ παραλληλόγραμμον, τοῦ δὲ ΔΓΖ διπλάσιον τὸ ΓΕ παραλληλόγραμμον· ἴσον ἄρα τὸ ΒΗ παραλληλόγραμμον τῷ ΓΕ παραλληλογράμμῳ.
And the parallelogram BG is double of ABH, and the parallelogram CE is double of DCF; therefore the parallelogram BG is equal to the parallelogram CE.
ὁμοίως δὴ δείξομεν, ὅτι καὶ τὸ μὲν ΑΓ τῷ ΗΖ ἐστιν ἴσον, τὸ δὲ ΑΕ τῷ ΒΖ.
ἐὰν ἄρα στερεὸν ὑπὸ παραλλήλων ἐπιπέδων περιέχηται, τὰ ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά ἐστιν· ὅπερ ἔδει δεῖξαι.
Similarly indeed we shall show that AC is also equal to GF, and AE to BF. Therefore, if a solid is contained by parallel planes, its opposite planes are equal and parallelograms; which was required to show.
§11.prop.25ἐὰν στερεὸν παραλληλεπίπεδον ἐπιπέδῳ τμηθῇ παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔσται ὡς ἡ βάσις πρὸς τὴν βάσιν, οὕτως τὸ στερεὸν πρὸς τὸ στερεόν.
If a parallelepipedal solid is cut by a plane which is parallel to the opposite planes, the base will be to the base as the solid is to the solid.
στερεὸν γὰρ παραλληλεπίπεδον τὸ ΑΒΓΔ ἐπιπέδῳ τῷ ΖΗ τετμήσθω παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις τοῖς ΡΑ, ΔΘ· λέγω, ὅτι ἐστὶν ὡς ἡ ΑΕΖΦ βάσις πρὸς τὴν ΕΘΓΖ βάσιν, οὕτως τὸ ΑΒΖΥ στερεὸν πρὸς τὸ ΕΗΓΔ στερεόν.
For let the parallelepipedal solid ABCD be cut by the plane FG which is parallel to the opposite planes RA, DH; I say that as the base AEFV is to the base EHCF, so is the solid ABFU to the solid EGCD.
Ἐκβεβλήσθω γὰρ ἡ ΑΘ ἐφʼ ἑκάτερα τὰ μέρη, καὶ κείσθωσαν τῇ μὲν ΑΕ ἴσαι ὁσαιδηποτοῦν αἱ ΑΚ, ΚΛ, τῇ δὲ ΕΘ ἴσαι ὁσαιδηποτοῦν αἱ ΘΜ, ΜΝ, καὶ συμπεπληρώσθω τὰ ΛΟ, ΚΦ, ΘΧ, ΜΣ παραλληλόγραμμα καὶ τὰ ΛΠ, ΚΡ, ΔΜ, ΜΤ στερεά.
For let AH be produced in each direction, and let there be set out any number of straight lines AK, KL equal to AE, and any number of straight lines HM, MN equal to EH, and let the parallelepipedons LO, KV, HW, MS and the solids LP, KR, DM, MT be completed.
καὶ ἐπεὶ ἴσαι εἰσὶν αἱ ΛΚ, ΚΑ, ΑΕ εὐθεῖαι ἀλλήλαις, ἴσα ἐστὶ καὶ τὰ μὲν ΛΟ, ΚΦ, ΑΖ παραλληλόγραμμα ἀλλήλοις, τὰ δὲ ΚΞ, ΚΒ, ΑΗ ἀλλήλοις καὶ ἔτι τὰ ΛΨ, ΚΠ, ΑΡ ἀλλήλοις· ἀπεναντίον γάρ.
And since the straight lines LK, KA, AE are equal to one another, the parallelograms LO, KV, AF are also equal to one another, and KX, KB, AG to one another, and further LY, KP, AR to one another; for they are opposite.
διὰ τὰ αὐτὰ δὴ καὶ τὰ μὲν ΕΓ, ΘΧ, ΜΣ παραλληλόγραμμα ἴσα εἰσὶν ἀλλήλοις, τὰ δὲ ΘΗ, ΘΙ, ΙΝ ἴσα εἰσὶν ἀλλήλοις, καὶ ἔτι τὰ ΔΘ, ΜΩ, ΝΤ· τρία ἄρα ἐπίπεδα τῶν ΛΠ, ΚΡ, ΑΥ στερεῶν τρισὶν ἐπιπέδοις ἐστὶν ἴσα.
For the same reasons indeed, the parallelograms EC, HW, MS are also equal to one another, and HG, HI, IN are equal to one another, and further DH, MZ, NT; therefore three planes of the solids LP, KR, AU are equal to three planes.
ἀλλὰ τὰ τρία τρισὶ τοῖς ἀπεναντίον ἐστὶν ἴσα· τὰ ἄρα τρία στερεὰ τὰ ΛΠ, ΚΡ, ΑΥ ἴσα ἀλλήλοις ἐστίν.
But the three are equal to the three opposite; therefore the three solids LP, KR, AU are equal to one another.
διὰ τὰ αὐτὰ δὴ καὶ τὰ τρία στερεὰ τὰ ΕΔ, ΔΜ, ΜΤ ἴσα ἀλλήλοις ἐστίν· ὁσαπλασίων ἄρα ἐστὶν ἡ ΛΖ βάσις τῆς ΑΖ βάσεως, τοσαυταπλάσιόν ἐστι καὶ τὸ ΛΥ στερεὸν τοῦ ΑΥ στερεοῦ.
For the same reasons indeed, the three solids ED, DM, MT are also equal to one another; therefore, whatever multiple the base LF is of the base AF, that same multiple is the solid LU of the solid AU.
διὰ τὰ αὐτὰ δὴ ὁσαπλασίων ἐστὶν ἡ ΝΖ βάσις τῆς ΖΘ βάσεως, τοσαυταπλάσιόν ἐστι καὶ τὸ ΝΥ στερεὸν τοῦ ΘΥ στερεοῦ.
For the same reasons indeed, whatever multiple the base NF is of the base FH, that same multiple is the solid NU of the solid HU.
καὶ εἰ ἴση ἐστὶν ἡ ΛΖ βάσις τῇ ΝΖ βάσει, ἴσον ἐστὶ καὶ τὸ ΛΥ στερεὸν τῷ ΝΥ στερεῷ, καὶ εἰ ὑπερέχει ἡ ΛΖ βάσις τῆς ΝΖ βάσεως, ὑπερέχει καὶ τὸ ΛΥ στερεὸν τοῦ ΝΥ στερεοῦ, καὶ εἰ ἐλλείπει, ἐλλείπει.
And if the base LF is equal to the base NF, the solid LU is also equal to the solid NU, and if the base LF exceeds the base NF, the solid LU also exceeds the solid NU, and if it falls short, it falls short.
τεσσάρων δὴ ὄντων μεγεθῶν, δύο μὲν βάσεων τῶν ΑΖ, ΖΘ, δύο δὲ στερεῶν τῶν ΑΥ, ΥΘ, εἴληπται ἰσάκις πολλαπλάσια τῆς μὲν ΑΖ βάσεως καὶ τοῦ ΑΥ στερεοῦ ἥ τε ΛΖ βάσις καὶ τὸ ΛΥ στερεόν, τῆς δὲ ΘΖ βάσεως καὶ τοῦ ΘΥ στερεοῦ ἥ τε ΝΖ βάσις καὶ τὸ ΝΥ στερεόν, καὶ δέδεικται, ὅτι εἰ ὑπερέχει ἡ ΛΖ βάσις τῆς ΖΝ βάσεως, ὑπερέχει καὶ τὸ ΛΥ στερεὸν τοῦ ΝΥ, καὶ εἰ ἴση, ἴσον, καὶ εἰ ἐλλείπει, ἐλλείπει.
Since then there are four magnitudes, namely two bases AF, FH, and two solids AU, HU, there have been taken equal multiples of the base AF and of the solid AU, namely the base LF and the solid LU, and of the base FH and of the solid HU, namely the base NF and the solid NU, and it has been shown that, if the base LF exceeds the base FN, the solid LU also exceeds the solid NU, and if equal, equal, and if it falls short, it falls short.
ἔστιν ἄρα ὡς ἡ ΑΖ βάσις πρὸς τὴν ΖΘ βάσιν, οὕτως τὸ ΑΥ στερεὸν πρὸς τὸ ΥΘ στερεόν· ὅπερ ἔδει δεῖξαι.
Therefore, as the base AF is to the base FH, so is the solid AU to the solid HU; which was required to show.