§12.prop.10#2λέγω δή, ὅτι οὐδὲ ἐλάττων ἐστὶν ἢ τριπλάσιος ὁ κύλινδρος τοῦ κώνου.
I say indeed that neither is the cylinder less than triple of the cone.
εἰ γὰρ δυνατόν, ἔστω ἐλάττων ἢ τριπλάσιος ὁ κύλινδρος τοῦ κώνου· ἀνάπαλιν ἄρα ὁ κῶνος τοῦ κυλίνδρου μείζων ἐστὶν ἢ τρίτον μέρος.
For, if possible, let the cylinder be less than triple of the cone; therefore, conversely, the cone is greater than a third part of the cylinder.
ἐγγεγράφθω δὴ εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον τὸ ΑΒΓΔ· τὸ ΑΒΓΔ ἄρα τετράγωνον μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ ΑΒΓΔ κύκλου.
Let there be inscribed indeed in the circle ABCD the square ABCD; therefore the square ABCD is greater than the half of the circle ABCD.
καὶ ἀνεστάτω ἀπὸ τοῦ ΑΒΓΔ τετραγώνου πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ κώνῳ· ἡ ἄρα ἀνασταθεῖσα πυραμὶς μείζων ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ κώνου, ἐπειδήπερ, ὡς ἔμπροσθεν ἐδείκνυμεν, ὅτι ἐὰν περὶ τὸν κύκλον τετράγωνον περιγράψωμεν, ἔσται τὸ ΑΒΓΔ τετράγωνον ἥμισυ τοῦ περὶ τὸν κύκλον περιγεγραμμένου τετραγώνου· καὶ ἐὰν ἀπὸ τῶν τετραγώνων στερεὰ παραλληλεπίπεδα ἀναστήσωμεν ἰσοϋψῆ τῷ κώνῳ, ἃ καὶ καλεῖται πρίσματα, ἔσται τὸ ἀνασταθὲν ἀπὸ τοῦ ΑΒΓΔ τετραγώνου ἥμισυ τοῦ ἀνασταθέντος ἀπὸ τοῦ περὶ τὸν κύκλον περιγραφέντος τετραγώνου·
And let there be set up from the square ABCD a pyramid having the same vertex as the cone; therefore the pyramid so set up is greater than the half part of the cone, because, as we proved before, that if we circumscribe a square about the circle, the square ABCD will be half of the square circumscribed about the circle; and if from the squares we set up parallelepipedal solids of equal height with the cone, which are also called prisms, the one set up from the square ABCD will be half of that set up from the square circumscribed about the circle; for they are to one another as their bases.
πρὸς ἄλληλα γάρ εἰσιν ὡς αἱ βάσεις. ὥστε καὶ τὰ τρίτα· καὶ πυραμὶς ἄρα, ἧς βάσις τὸ ΑΒΓΔ τετράγωνον, ἥμισύ ἐστι τῆς πυραμίδος τῆς ἀνασταθείσης ἀπὸ τοῦ περὶ τὸν κύκλον περιγραφέντος τετραγώνου.
So that also their third parts; therefore also the pyramid whose base is the square ABCD is half of the pyramid set up from the square circumscribed about the circle.
καί ἐστι μείζων ἡ πυραμὶς ἡ ἀνασταθεῖσα ἀπὸ τοῦ περὶ τὸν κύκλον τετραγώνου τοῦ κώνου· ἐμπεριέχει γὰρ αὐτόν.
And the pyramid set up from the square circumscribed about the circle is greater than the cone; for it contains it.
ἡ ἄρα πυραμὶς, ἧς βάσις τὸ ΑΒΓΔ τετράγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, μείζων ἐστὶν ἢ τὸ ἥμισυ τοῦ κώνου.
Therefore the pyramid whose base is the square ABCD and vertex the same as the cone is greater than the half of the cone.
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 μέρος τοῦ καθʼ ἑαυτὸ τμήματος τοῦ ΑΒΓΔ κύκλου.
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 part of the segment of the circle ABCD corresponding to it.
καὶ ἀνεστάτωσαν ἐφʼ ἑκάστου τῶν ΑΕΒ, ΒΖΓ, ΓΗΔ, ΔΘΑ τριγώνων πυραμίδες τὴν αὐτὴν κορυφὴν ἔχουσαι τῷ κώνῳ· καὶ ἑκάστη ἄρα τῶν ἀνασταθεισῶν πυραμίδων κατὰ τὸν αὐτὸν τρόπον μείζων ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ καθʼ ἑαυτὴν τμήματος τοῦ κώνου.
And let there be set up on each of the triangles AEB, BZG, GHD, DThA pyramids having the same vertex as the cone; therefore also each of the pyramids set up is in the same manner greater than the half part of the segment of the cone corresponding to it.
τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ ἀνιστάντες ἐφʼ ἑκάστου τῶν τριγώνων πυραμίδα τὴν αὐτὴν κορυφὴν ἔχουσαν τῷ κώνῳ καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κώνου, ἃ ἔσται ἐλάττονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ κῶνος τοῦ τρίτου μέρους τοῦ κυλίνδρου.
Therefore, bisecting the remaining circumferences and joining straight lines and setting up on each of the triangles a pyramid having the same vertex as the cone and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone exceeds the third part of the cylinder.
λελείφθω, καὶ ἔστω τὰ ἐπὶ τῶν ΑΕ, ΕΒ, ΒΖ, ΖΓ, ΓΗ, ΗΔ, ΔΘ, ΘΑ· λοιπὴ ἄρα ἡ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, μείζων ἐστὶν ἢ τρίτον μέρος τοῦ κυλίνδρου.
Let them be left, and let them be those on AE, EB, BZ, ZG, GH, HD, DTh, ThA; therefore, the remaining pyramid, whose base is the polygon AEBZGHDTh and vertex the same as the cone, is greater than a third part of the cylinder.
ἀλλʼ ἡ πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΕΒΖΓ ΗΔΘ πολύγωνον, κορυφὴ δὲ ἡ αὐτὴ τῷ κώνῳ, τρίτον ἐστὶ μέρος τοῦ πρίσματος, οὗ βάσις μέν ἐστι τὸ ΑΕΒΖΓ ΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ· τὸ ἄρα πρίσμα, οὗ βάσις μέν ἐστι τὸ ΑΕΒΖΓΗΔΘ πολύγωνον, ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, μεῖζόν ἐστι τοῦ κυλίνδρου, οὗ βάσις ἐστὶν ὁ ΑΒΓΔ κύκλος.
But the pyramid, whose base is the polygon AEBZGHDTh and vertex the same as the cone, is a third part of the prism whose base is the polygon AEBZG HDTh and height the same as the cylinder; therefore the prism, whose base is the polygon AEBZGHDTh and height the same as the cylinder, is greater than the cylinder whose base is the circle ABCD.
ἀλλὰ καὶ ἔλαττον· ἐμπεριέχεται γὰρ ὑπʼ αὐτοῦ· ὅπερ ἐστὶν ἀδύνατον.
But it is also less; for it is contained by it; which is impossible.
οὐκ ἄρα ὁ κύλινδρος τοῦ κώνου ἐλάττων ἐστὶν ἢ τριπλάσιος.
Therefore the cylinder is not less than triple of the cone.
ἐδείχθη δέ, ὅτι οὐδὲ μείζων ἢ τριπλάσιος· τριπλάσιος ἄρα ὁ κύλινδρος τοῦ κώνου· ὥστε ὁ κῶνος τρίτον ἐστὶ μέρος τοῦ κυλίνδρου.
And it was proved that neither is it greater than triple; therefore the cylinder is triple of the cone; so that the cone is a third part of the cylinder.
πᾶς ἄρα κῶνος κυλίνδρου τρίτον μέρος ἐστὶ τοῦ τὴν αὐτὴν βάσιν ἔχοντος αὐτῷ καὶ ὕψος ἴσον· ὅπερ ἔδει δεῖξαι.
Therefore every cone is a third part of the cylinder which has the same base with it and equal height; which it was required to prove.