§12.prop.13ἐὰν κύλινδρος ἐπιπέδῳ τμηθῇ παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔσται ὡς ὁ κύλινδρος πρὸς τὸν κύλινδρον, οὕτως ὁ ἄξων πρὸς τὸν ἄξονα.
If a cylinder is cut by a plane parallel to the opposite faces, as the cylinder is to the cylinder, so will the axis be to the axis.
κύλινδρος γὰρ ὁ ΑΔ ἐπιπέδῳ τῷ ΗΘ τετμήσθω παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις τοῖς ΑΒ, ΓΔ, καὶ συμβαλλέτω τῷ ἄξονι τὸ ΗΘ ἐπίπεδον κατὰ τὸ Κ σημεῖον· λέγω, ὅτι ἐστὶν ὡς ὁ ΒΗ κύλινδρος πρὸς τὸν ΗΔ κύλινδρον, οὕτως ὁ ΕΚ ἄξων πρὸς τὸν ΚΖ ἄξονα.
For let the cylinder AD be cut by the plane HTh parallel to the opposite faces AB, GD, and let the plane HTh meet the axis at the point K; I say that, as the cylinder BH is to the cylinder HD, so is the axis EK to the axis KZ.
Ἐκβεβλήσθω γὰρ ὁ ΕΖ ἄξων ἐφʼ ἑκάτερα τὰ μέρη ἐπὶ τὰ Λ, Μ σημεῖα, καὶ ἐκκείσθωσαν τῷ ΕΚ ἄξονι ἴσοι ὁσοιδηποτοῦν οἱ ΕΝ, ΝΛ, τῷ δὲ ΖΚ ἴσοι ὁσοιδηποτοῦν οἱ ΖΞ, ΞΜ, καὶ νοείσθω ὁ ἐπὶ τοῦ ΛΜ ἄξονος κύλινδρος ὁ ΟΧ, οὗ βάσεις οἱ ΟΠ, ΦΧ κύκλοι.
For let the axis EZ be produced in both directions to the points L, M, and let any number of segments EN, NL be set out equal to the axis EK, and any number of segments ZX, XM equal to ZK, and let the cylinder OX on the axis LM be conceived, of which the bases are the circles OP, PhCh.
καὶ ἐκβεβλήσθω διὰ τῶν Ν, Ξ σημείων ἐπίπεδα παράλληλα τοῖς ΑΒ, ΓΔ καὶ ταῖς βάσεσι τοῦ ΟΧ κυλίνδρου καὶ ποιείτωσαν τοὺς ΡΣ, ΤΥ κύκλους περὶ τὰ Ν, Ξ κέντρα.
And let planes parallel to AB, GD and to the bases of the cylinder OX be produced through the points N, X, and let them make the circles RS, TY about the centers N, X.
καὶ ἐπεὶ οἱ ΛΝ, ΝΕ, ΕΚ ἄξονες ἴσοι εἰσὶν ἀλλήλοις, οἱ ἄρα ΠΡ, ΡΒ, ΒΗ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις.
And since the axes LN, NE, EK are equal to one another, therefore the cylinders PR, RB, BH are to one another as their bases.
ἴσαι δέ εἰσιν αἱ βάσεις· ἴσοι ἄρα καὶ οἱ ΠΡ, ΡΒ, ΒΗ κύλινδροι ἀλλήλοις.
But the bases are equal; therefore the cylinders PR, RB, BH are also equal to one another.
ἐπεὶ οὖν οἱ ΛΝ, ΝΕ, ΕΚ ἄξονες ἴσοι εἰσὶν ἀλλήλοις, εἰσὶ δὲ καὶ οἱ ΠΡ, ΡΒ, ΒΗ κύλινδροι ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῷ πλήθει, ὁσαπλασίων ἄρα ὁ ΚΛ ἄξων τοῦ ΕΚ ἄξονος, τοσαυταπλασίων ἔσται καὶ ὁ ΠΗ κύλινδρος τοῦ ΗΒ κυλίνδρου.
Since, therefore, the axes LN, NE, EK are equal to one another, and the cylinders PR, RB, BH are also equal to one another, and the number is equal to the number, therefore, of whatever multiple the axis KL is of the axis EK, of the same multiple will the cylinder PH also be of the cylinder HB.
διὰ τὰ αὐτὰ δὴ καὶ ὁσαπλασίων ἐστὶν ὁ ΜΚ ἄξων τοῦ ΚΖ ἄξονος, τοσαυταπλασίων ἐστὶ καὶ ὁ ΧΗ κύλινδρος τοῦ ΗΔ κυλίνδρου.
For the same reasons also, of whatever multiple the axis MK is of the axis KZ, of the same multiple is the cylinder ChH also of the cylinder HD.
καὶ εἰ μὲν ἴσος ἐστὶν ὁ ΚΛ ἄξων τῷ ΚΜ ἄξονι, ἴσος ἔσται καὶ ὁ ΠΗ κύλινδρος τῷ ΗΧ κυλίνδρῳ, εἰ δὲ μείζων ὁ ἄξων τοῦ ἄξονος, μείζων καὶ ὁ κύλινδρος τοῦ κυλίνδρου, καὶ εἰ ἐλάσσων, ἐλάσσων.
And if the axis KL is equal to the axis KM, the cylinder PH will also be equal to the cylinder HX, and if the axis is greater than the axis, the cylinder will also be greater than the cylinder, and if less, less.
τεσσάρων δὴ μεγεθῶν ὄντων, ἀξόνων μὲν τῶν ΕΚ, ΚΖ, κυλίνδρων δὲ τῶν ΒΗ, ΗΔ, εἴληπται ἰσάκις πολλαπλάσια, τοῦ μὲν ΕΚ ἄξονος καὶ τοῦ ΒΗ κυλίνδρου ὅ τε ΛΚ ἄξων καὶ ὁ ΠΗ κύλινδρος, τοῦ δὲ ΚΖ ἄξονος καὶ τοῦ ΗΔ κυλίνδρου ὅ τε ΚΜ ἄξων καὶ ὁ ΗΧ κύλινδρος, καὶ δέδεικται, ὅτι εἰ ὑπερέχει ὁ ΚΛ ἄξων τοῦ ΚΜ ἄξονος, ὑπερέχει καὶ ὁ ΠΗ κύλινδρος τοῦ ΗΧ κυλίνδρου, καὶ εἰ ἴσος, ἴσος, καὶ εἰ ἐλάσσων, ἐλάσσων.
Thus, there being four magnitudes, namely the axes EK, KZ and the cylinders BH, HD, equimultiples have been taken, of the axis EK and the cylinder BH, namely the axis LK and the cylinder PH, and of the axis KZ and the cylinder HD, namely the axis KM and the cylinder HX, and it has been shown that, if the axis KL exceeds the axis KM, the cylinder PH also exceeds the cylinder HX, and if equal, equal, and if less, less.
ἔστιν ἄρα ὡς ὁ ΕΚ ἄξων πρὸς τὸν ΚΖ ἄξονα, οὕτως ὁ ΒΗ κύλινδρος πρὸς τὸν ΗΔ κύλινδρον· ὅπερ ἔδει δεῖξαι.
Therefore, as the axis EK is to the axis KZ, so is the cylinder BH to the cylinder HD; which it was required to show.
§12.prop.14οἱ ἐπὶ ἴσων βάσεων ὄντες κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ὕψη.
Cones and cylinders which are on equal bases are to one another as their heights.
ἔστωσαν γὰρ ἐπὶ ἴσων βάσεων τῶν ΑΒ, ΓΔ κύκλων κύλινδροι οἱ ΕΒ, ΖΔ· λέγω, ὅτι ἐστὶν ὡς ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΗΘ ἄξων πρὸς τὸν ΚΛ ἄξονα.
For let there be cylinders EB, ZD on equal bases, namely the circles AB, GD; I say that, as the cylinder EB is to the cylinder ZD, so is the axis HTh to the axis KL.
Ἐκβεβλήσθω γὰρ ὁ ΚΛ ἄξων ἐπὶ τὸ Ν σημεῖον, καὶ κείσθω τῷ ΗΘ ἄξονι ἴσος ὁ ΛΝ, καὶ περὶ ἄξονα τὸν ΛΝ κύλινδρος νενοήσθω ὁ ΓΜ. ἐπεὶ οὖν οἱ ΕΒ, ΓΜ κύλινδροι ὑπὸ τὸ αὐτὸ ὕψος εἰσίν, πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις.
For let the axis KL be produced to the point N, and let LN be set out equal to the axis HTh, and let the cylinder GM be conceived about the axis LN. Since, therefore, the cylinders EB, GM are under the same height, they are to one another as their bases.
ἴσαι δέ εἰσιν αἱ βάσεις ἀλλήλαις· ἴσοι ἄρα εἰσὶ καὶ οἱ ΕΒ, ΓΜ κύλινδροι.
But the bases are equal to one another; therefore the cylinders EB, GM are also equal.
καὶ ἐπεὶ κύλινδρος ὁ ΖΜ ἐπιπέδῳ τέτμηται τῷ ΓΔ παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔστιν ἄρα ὡς ὁ ΓΜ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΛΝ ἄξων πρὸς τὸν ΚΛ ἄξονα.
And since the cylinder ZM is cut by the plane GD which is parallel to the opposite faces, therefore, as the cylinder GM is to the cylinder ZD, so is the axis LN to the axis KL.
ἴσος δέ ἐστιν ὁ μὲν ΓΜ κύλινδρος τῷ ΕΒ κυλίνδρῳ, ὁ δὲ ΛΝ ἄξων τῷ ΗΘ ἄξονι· ἔστιν ἄρα ὡς ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΗΘ ἄξων πρὸς τὸν ΚΛ ἄξονα.
But the cylinder GM is equal to the cylinder EB, and the axis LN to the axis HTh; therefore, as the cylinder EB is to the cylinder ZD, so is the axis HTh to the axis KL.
ὡς δὲ ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΑΒΗ κῶνος πρὸς τὸν ΓΔΚ κῶνον.
But as the cylinder EB is to the cylinder ZD, so is the cone ABH to the cone GDK.
καὶ ὡς ἄρα ὁ ΗΘ ἄξων πρὸς τὸν ΚΛ ἄξονα, οὕτως ὁ ΑΒΗ κῶνος πρὸς τὸν ΓΔΚ κῶνον καὶ ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον· ὅπερ ἔδει δεῖξαι.
Therefore also, as the axis HTh is to the axis KL, so is the cone ABH to the cone GDK and the cylinder EB to the cylinder ZD; which it was required to show.