§12.prop.10#1πᾶς κῶνος κυλίνδρου τρίτον μέρος ἐστὶ τοῦ τὴν αὐτὴν βάσιν ἔχοντος αὐτῷ καὶ ὕψος ἴσον.
Every cone is a third part of the cylinder which has the same base with it and equal height.
ἐχέτω γὰρ κῶνος κυλίνδρῳ βάσιν τε τὴν αὐτὴν τὸν ΑΒΓΔ κύκλον καὶ ὕψος ἴσον·
For let a cone have the same base, the circle ABCD, and equal height with a cylinder.
λέγω, ὅτι ὁ κῶνος τοῦ κυλίνδρου τρίτον ἐστὶ μέρος, τουτέστιν ὅτι ὁ κύλινδρος τοῦ κώνου τριπλασίων ἐστίν.
I say that the cone is a third part of the cylinder, that is, that the cylinder is triple of the cone.
εἰ γὰρ μή ἐστιν ὁ κύλινδρος τοῦ κώνου τριπλασίων, ἔσται ὁ κύλινδρος τοῦ κώνου ἤτοι μείζων ἢ τριπλασίων ἢ ἐλάσσων ἢ τριπλασίων.
For if the cylinder is not triple of the cone, the cylinder will be either greater than triple or less than triple.
ἔστω πρότερον μείζων ἢ τριπλασίων, καὶ ἐγγεγράφθω εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον τὸ ΑΒΓΔ· τὸ δὴ ΑΒΓΔ τετράγωνον μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ ΑΒΓΔ κύκλου.
Let it first be greater than triple, and let the square ABCD be inscribed in the circle ABCD; the square ABCD is indeed greater than the half of the circle ABCD.
καὶ ἀνεστάτω ἀπὸ τοῦ ΑΒΓΔ τετραγώνου πρίσμα ἰσουψὲς τῷ κυλίνδρῳ.
And let there be set up from the square ABCD a prism of equal height with the cylinder.
τὸ δὴ ἀνιστάμενον πρίσμα μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ κυλίνδρου, ἐπειδήπερ κἂν περὶ τὸν ΑΒΓΔ κύκλον τετράγωνον περιγράψωμεν, τὸ ἐγγεγραμμένον εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον ἥμισύ ἐστι τοῦ περιγεγραμμένου· καί ἐστι τὰ ἀπʼ αὐτῶν ἀνιστάμενα στερεὰ παραλληλεπίπεδα πρίσματα ἰσοϋψῆ· τὰ δὲ ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις· καὶ τὸ ἐπὶ τοῦ ΑΒΓΔ ἄρα τετραγώνου ἀνασταθὲν πρίσμα ἥμισύ ἐστι τοῦ ἀνασταθέντος πρίσματος ἀπὸ τοῦ περὶ τὸν ΑΒΓΔ κύκλον περιγραφέντος τετραγώνου· καί ἐστιν ὁ κύλινδρος ἐλάττων τοῦ πρίσματος τοῦ ἀνασταθέντος ἀπὸ τοῦ περὶ τὸν ΑΒΓΔ κύκλον περιγραφέντος τετραγώνου· τὸ ἄρα πρίσμα τὸ ἀνασταθὲν ἀπὸ τοῦ ΑΒΓΔ τετραγώνου ἰσοϋψὲς τῷ κυλίνδρῳ μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ κυλίνδρου.
The prism so set up is indeed greater than the half of the cylinder, because even if we circumscribe a square about the circle ABCD, the square inscribed in the circle ABCD is half of the circumscribed square; and the parallelepipedal solids of equal height set up from them, since the parallelepipedal solids which are under the same height are to one another as their bases; therefore also the prism set up on the square ABCD is half of the prism set up from the square circumscribed about the circle ABCD; and the cylinder is less than the prism set up from the square circumscribed about the circle ABCD; therefore the prism set up from the square ABCD of equal height with the cylinder is greater than the half of the cylinder.
τετμήσθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΑ περιφέρειαι δίχα κατὰ τὰ Ε, Ζ, Η, Θ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ· καὶ ἕκαστον ἄρα τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ καθʼ ἑαυτὸ τμήματος τοῦ ΑΒΓΔ κύκλου, ὡς ἔμπροσθεν ἐδείκνυμεν.
Let the circumferences AB, BC, CD, DA be bisected at the points E, Z, H, Th, and let AE, EB, BZ, ZG, GH, HD, DTh, ThA be joined; therefore also each of the triangles AEB, BZG, GHD, DThA is greater than the half of the segment of the circle ABCD corresponding to it, as we proved before.
ἀνεστάτω ἐφʼ ἑκάστου τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων πρίσματα ἰσουψῆ τῷ κυλίνδρῳ· καὶ ἕκαστον ἄρα τῶν ἀνασταθέντων πρισμάτων μεῖζόν ἐστιν ἢ τὸ ἥμισυ μέρος τοῦ καθʼ ἑαυτὸ τμήματος τοῦ κυλίνδρου, ἐπειδήπερ ἐὰν διὰ τῶν Ε, Ζ, Η, θ σημείων παραλλήλους ταῖς ΑΒ, ΒΓ, ΓΔ, ΔΑ ἀγάγωμεν, καὶ συμπληρώσωμεν τὰ ἐπὶ τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ παραλληλόγραμμα, καὶ ἀπʼ αὐτῶν ἀναστήσωμεν στερεὰ παραλληλεπίπεδα ἰσοϋψῆ τῷ κυλίνδρῳ, ἑκάστου τῶν ἀνασταθέντων ἡμίση ἐστὶ τὰ πρίσματα τὰ ἐπὶ τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων· καί ἐστι τὰ τοῦ κυλίνδρου τμήματα ἐλάττονα τῶν ἀνασταθέντων στερεῶν παραλληλεπιπέδων· ὥστε καὶ τὰ ἐπὶ τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῶν καθʼ ἑαυτὰ τοῦ κυλίνδρου τμημάτων.
Let there be set up on each of the triangles AEB, BZG, GHD, DThA prisms of equal height with the cylinder; therefore also each of the prisms set up is greater than the half part of the segment of the cylinder corresponding to it, because if through the points E, Z, H, Th we draw parallels to AB, BC, CD, DA, and complete the parallelograms on AB, BC, CD, DA, and set up from them parallelepipedal solids of equal height with the cylinder, the prisms on the triangles AEB, BZG, GHD, DThA are halves of each of those set up; and the segments of the cylinder are less than the parallelepipedal solids set up; so that also the prisms on the triangles AEB, BZG, GHD, DThA are greater than the half of the segments of the cylinder corresponding to them.
τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐφʼ ἑκάστου τῶν τριγώνων πρίσματα ἰσοϋψῆ τῷ κυλίνδρῳ καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κυλίνδρου, ἃ ἔσται ἐλάττονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ κύλινδρος τοῦ τριπλασίου τοῦ κώνου.
Therefore, bisecting the remaining circumferences and joining straight lines and setting up on each of the triangles prisms of equal height with the cylinder and doing this continually, we shall leave some segments of the cylinder which will be less than the excess by which the cylinder exceeds triple of the cone.
λελείφθω, καὶ ἔστω τὰ ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ· λοιπὸν ἄρα τὸ πρίσμα, οὗ βάσις μὲν τὸ ΑΕΒΖ ΓΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, μεῖζόν ἐστιν ἢ τριπλάσιον τοῦ κώνου.
Let them be left, and let them be AE, EB, BZ, ZG, GH, HD, DTh, ThA; therefore, the remaining prism, whose base is the polygon AEBZGHDTh and height the same as the cylinder, is greater than triple of the cone.
ἀλλὰ τὸ πρίσμα, οὗ βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, τριπλάσιόν ἐστι τῆς πυραμίδος, ἧς βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ· καὶ ἡ πυραμὶς ἄρα, ἧς βάσις μὲν τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, μείζων ἐστὶ τοῦ κώνου τοῦ βάσιν ἔχοντος τὸν ΑΒ ΓΔ κύκλον.
But the prism, whose base is the polygon AEBZGHDTh and height the same as the cylinder, is triple of the pyramid whose base is the polygon AEBZGHDTh and vertex the same as the cone; therefore also the pyramid whose base is the polygon AEBZGHDTh and vertex the same as the cone is greater than the cone which has the circle AB CD as base.
ἀλλὰ καὶ ἐλάττων· ἐμπεριέχεται γὰρ ὑπʼ αὐτοῦ· ὅπερ ἐστὶν ἀδύνατον.
But it is also less; for it is contained by it; which is impossible.
οὐκ ἄρα ἐστὶν ὁ κύλινδρος τοῦ κώνου μείζων ἢ τριπλάσιος.
Therefore the cylinder is not greater than triple of the cone.