14. Cones and cylinders which are on equal bases are to one another as their heights. For let EB, FD be cylinders on equal bases, the circles AB, CD; I say that, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL. For let the axis KL be produced to the point N, let LN be made equal to the axis GH, and let the cylinder CM be conceived about LN as axis. Since then the cylinders EB, CM are of the same height, they are to one another as their bases. [] But the bases are equal to one another; therefore the cylinders EB, CM are also equal. And, since the cylinder FM has been cut by the plane CD which is parallel to its opposite planes, therefore, as the cylinder CM is to the cylinder FD, so is the axis LN to the axis KL. [] But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore, as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL. But, as the cylinder EB is to the cylinder FD, so is the cone ABG to the cone CDK. [] Therefore also, as the axis GH is to the axis KL, so is the cone ABG to the cone CDK and the cylinder EB to the cylinder FD. Q. E. D.
14. οἱ ἐπὶ ἴσων βάσεων ὄντες κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ὕψη. 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. 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. ἴσαι δέ εἰσιν αἱ βάσεις ἀλλήλαις· ἴσοι ἄρα εἰσὶ καὶ οἱ ΕΒ, ΓΜ κύλινδροι. 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. 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. ὡς δὲ ὁ ΕΒ κύλινδρος πρὸς τὸν ΖΔ κύλινδρον, οὕτως ὁ ΑΒΗ κῶνος πρὸς τὸν ΓΔΚ κῶνον. 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.
15. In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal. Let there be equal cones and cylinders of which the circles ABCD, EFGH are the bases; let AC, EG be the diameters of the bases, and KL, MN the axes, which are also the heights of the cones or cylinders; let the cylinders AO, EP be completed. I say that in the cylinders AO, EP the bases are reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so is the height MN to the height KL. For the height LK is either equal to the height MN or not equal. First, let it be equal. Now the cylinder AO is also equal to the cylinder EP. But cones and cylinders which are of the same height are to one another as their bases; [] therefore the base ABCD is also equal to the base EFGH. Hence also, reciprocally, as the base ABCD is to the base EFGH, so is the height MN to the height KL. Next, let the height LK not be equal to MN, but let MN be greater; from the height MN let QN be cut off equal to KL, through the point Q let the cylinder EP be cut by the plane TUS parallel to the planes of the circles EFGH, RP, and let the cylinder ES be conceived erected from the circle EFGH as base and with height NQ. Now, since the cylinder AO is equal to the cylinder EP, therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [] But, as the cylinder AO is to the cylinder ES, so is the base ABCD to the base EFGH, for the cylinders AO, ES are of the same height; [] and, as the cylinder EP is to the cylinder ES, so is the height MN to the height QN, for the cylinder EP has been cut by a plane which is parallel to its opposite planes. [] Therefore also, as the base ABCD is to the base EFGH, so is the height MN to the height QN. [] But the height QN is equal to the height KL; therefore, as the base ABCD is to the base EFGH, so is the height MN to the height KL. Therefore in the cylinders AO, EP the bases are reciprocally proportional to the heights. Next, in the cylinders AO, EP let the bases be reciprocally proportional to the heights, that is, as the base ABCD is to the base EFGH, so let the height MN be to the height KL; I say that the cylinder AO is equal to the cylinder EP. For, with the same construction, since, as the base ABCD is to the base EFGH, so is the height MN to the height KL, while the height KL is equal to the height QN, therefore, as the base ABCD is to the base EFGH, so is the height MN to the height QN But, as the base ABCD is to the base EFGH, so is the cylinder AO to the cylinder ES, for they are of the same height; [] and, as the height MN is to QN, so is the cylinder EP to the cylinder ES; [] therefore, as the cylinder AO is to the cylinder ES, so is the cylinder EP to the cylinder ES. [] Therefore the cylinder AO is equal to the cylinder EP. [] And the same is true for the cones also. Q. E. D.
15. τῶν ἴσων κώνων καὶ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν· καὶ ὧν κώνων καὶ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, ἴσοι εἰσὶν ἐκεῖνοι. ἔστωσαν ἴσοι κῶνοι καὶ κύλινδροι, ὧν βάσεις μὲν οἱ ΑΒΓΔ, ΕΖΗΘ κύκλοι, διάμετροι δὲ αὐτῶν αἱ ΑΓ, ΕΗ, ἄξονες δὲ οἱ ΚΛ, ΜΝ, οἵτινες καὶ ὕψη εἰσὶ τῶν κώνων ἢ κυλίνδρων, καὶ συμπεπληρώσθωσαν οἱ ΑΞ, ΕΟ κύλινδροι. λέγω, ὅτι τῶν ΑΞ, ΕΟ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν, καί ἐστιν ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος. τὸ γὰρ ΛΚ ὕψος τῷ ΜΝ ὕψει ἤτοι ἴσον ἐστὶν ἢ οὔ. ἔστω πρότερον ἴσον. ἔστι δὲ καὶ ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ ἴσος. οἱ δὲ ὑπὸ τὸ αὐτὸ ὕψος ὄντες κῶνοι καὶ κύλινδροι πρὸς ἀλλήλους εἰσὶν ὡς αἱ βάσεις· ἴση ἄρα καὶ ἡ ΑΒΓΔ βάσις τῇ ΕΖΗΘ βάσει. ὥστε καὶ ἀντιπέπονθεν, ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος. ἀλλὰ δὴ μὴ ἔστω τὸ ΛΚ ὕψος τῷ ΜΝ ἴσον, ἀλλʼ ἔστω μεῖζον τὸ ΜΝ, καὶ ἀφῃρήσθω ἀπὸ τοῦ ΜΝ ὕψους τῷ ΚΛ ἴσον τὸ ΠΝ, καὶ διὰ τοῦ Π σημείου τετμήσθω ὁ ΕΟ κύλινδρος ἐπιπέδῳ τῷ ΤΥΣ παραλλήλῳ τοῖς τῶν ΕΖΗΘ, ΡΟ κύκλων ἐπιπέδοις, καὶ ἀπὸ βάσεως μὲν τοῦ ΕΖΗΘ κύκλου, ὕψους δὲ τοῦ ΝΠ κύλινδρος νενοήσθω ὁ ΕΣ. καὶ ἐπεὶ ἴσος ἐστὶν ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ, ἔστιν ἄρα ὡς ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον, οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον. ἀλλʼ ὡς μὲν ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον, οὕτως ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ· ὑπὸ γὰρ τὸ αὐτὸ ὕψος εἰσὶν οἱ ΑΞ, ΕΣ κύλινδροι· ὡς δὲ ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος· ὁ γὰρ ΕΟ κύλινδρος ἐπιπέδῳ τέτμηται παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις. ἔστιν ἄρα καὶ ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος. ἴσον δὲ τὸ ΠΝ ὕψος τῷ ΚΛ ὕψει· ἔστιν ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος. τῶν ἄρα ΑΞ, ΕΟ κυλίνδρων ἀντιπεπόνθασιν αἱ βάσεις τοῖς ὕψεσιν. ἀλλὰ δὴ τῶν ΑΞ, ΕΟ κυλίνδρων ἀντιπεπονθέτωσαν αἱ βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος· λέγω, ὅτι ἴσος ἐστὶν ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ. τῶν γὰρ αὐτῶν κατασκευασθέντων ἐπεί ἐστιν ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος, ἴσον δὲ τὸ ΚΛ ὕψος τῷ ΠΝ ὕψει, ἔστιν ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος. ἀλλʼ ὡς μὲν ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον· ὑπὸ γὰρ τὸ αὐτὸ ὕψος εἰσίν· ὡς δὲ τὸ ΜΝ ὕψος πρὸς τὸ ΠΝ [ὕψος], οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον· ἔστιν ἄρα ὡς ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ κύλινδρον, οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ. ἴσος ἄρα ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ. ὡσαύτως δὲ καὶ ἐπὶ τῶν κώνων· ὅπερ ἔδει δεῖξαι.
16. Given two circles about the same centre, to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle. Let ABCD, EFGH be the two given circles about the same centre K; thus it is required to inscribe in the greater circle ABCD an equilateral polygon with an even number of sides which does not touch the circle EFGH. For let the straight line BKD be drawn through the centre K, and from the point G let GA be drawn at right angles to the straight line BD and carried through to C; therefore AC touches the circle EFGH. [] Then, bisecting the circumference BAD, bisecting the half of it, and doing this continually, we shall leave a circumference less than AD. [] Let such be left, and let it be LD; from L let LM be drawn perpendicular to BD and carried through to N, and let LD, DN be joined; therefore LD is equal to DN. [, ] Now, since LN is parallel to AC, and AC touches the circle EFGH, therefore LN does not touch the circle EFGH; therefore LD, DN are far from touching the circle EFGH. If then we fit into the circle ABCD straight lines equal to the straight line LD and placed continuously, there will be inscribed in the circle ABCD an equilateral polygon with an even number of sides which does not touch the lesser circle EFGH. Q. E. F.
16. δύο κύκλων περὶ τὸ αὐτὸ κέντρον ὄντων εἰς τὸν μείζονα κύκλον πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον ἐγγράψαι μὴ ψαῦον τοῦ ἐλάσσονος κύκλου. ἔστωσαν οἱ δοθέντες δύο κύκλοι οἱ ΑΒΓΔ, ΕΖΗΘ περὶ τὸ αὐτὸ κέντρον τὸ Κ· δεῖ δὴ εἰς τὸν μείζονα κύκλον τὸν ΑΒΓΔ πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον ἐγγράψαι μὴ ψαῦον τοῦ ΕΖΗΘ κύκλου. ἤχθω γὰρ διὰ τοῦ Κ κέντρου εὐθεῖα ἡ ΒΚΔ, καὶ ἀπὸ τοῦ Η σημείου τῇ ΒΔ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ ΗΑ καὶ διήχθω ἐπὶ τὸ Γ· ἡ ΑΓ ἄρα ἐφάπτεται τοῦ ΕΖΗΘ κύκλου. τέμνοντες δὴ τὴν ΒΑΔ περιφέρειαν δίχα καὶ τὴν ἡμίσειαν αὐτῆς δίχα καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομεν περιφέρειαν ἐλάσσονα τῆς ΑΔ. λελείφθω, καὶ ἔστω ἡ ΛΔ, καὶ ἀπὸ τοῦ Λ ἐπὶ τὴν ΒΔ κάθετος ἤχθω ἡ ΛΜ καὶ διήχθω ἐπὶ τὸ Ν, καὶ ἐπεζεύχθωσαν αἱ ΛΔ, ΔΝ· ἴση ἄρα ἐστὶν ἡ ΛΔ τῇ ΔΝ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΛΝ τῇ ΑΓ, ἡ δὲ ΑΓ ἐφάπτεται τοῦ ΕΖΗΘ κύκλου, ἡ ΛΝ ἄρα οὐκ ἐφάπτεται τοῦ ΕΖΗΘ κύκλου· πολλῷ ἄρα αἱ ΛΔ, ΔΝ οὐκ ἐφάπτονται τοῦ ΕΖΗΘ κύκλου. ἐὰν δὴ τῇ ΛΔ εὐθείᾳ ἴσας κατὰ τὸ συνεχὲς ἐναρμόσωμεν εἰς τὸν ΑΒΓΔ κύκλον, ἐγγραφήσεται εἰς τὸν ΑΒΓΔ κύκλον πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον μὴ ψαῦον τοῦ ἐλάσσονος κύκλου τοῦ ΕΖΗΘ· ὅπερ ἔδει ποιῆσαι.
17. Given two spheres about the same centre, to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.
17. δύο σφαιρῶν περὶ τὸ αὐτὸ κέντρον οὐσῶν εἰς τὴν μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγράψαι μὴ ψαῦον τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν.
Let two spheres be conceived about the same centre A; thus it is required to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.
Νενοήσθωσαν δύο σφαῖραι περὶ τὸ αὐτὸ κέντρον τὸ Α· δεῖ δὴ εἰς τὴν μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγράψαι μὴ ψαῦον τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν.
Let the spheres be cut by any plane through the centre; then the sections will be circles, inasmuch as the sphere was produced by the diameter remaining fixed and the semicircle being carried round it; [] hence, in whatever position we conceive the semicircle to be, the plane carried through it will produce a circle on the circumference of the sphere.
τετμήσθωσαν αἱ σφαῖραι ἐπιπέδῳ τινὶ διὰ τοῦ κέντρου· ἔσονται δὴ αἱ τομαὶ κύκλοι, ἐπειδήπερ μενούσης τῆς διαμέτρου καὶ περιφερομένου τοῦ ἡμικυκλίου ἐγίγνετο ἡ σφαῖρα· ὥστε καὶ καθʼ οἵας ἂν θέσεως ἐπινοήσωμεν τὸ ἡμικύκλιον, τὸ διʼ αὐτοῦ ἐκβαλλόμενον ἐπίπεδον ποιήσει ἐπὶ τῆς ἐπιφανείας τῆς σφαίρας κύκλον.
And it is manifest that this circle is the greatest possible, inasmuch as the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere.
καὶ φανερόν, ὅτι καὶ μέγιστον, ἐπειδήπερ ἡ διάμετρος τῆς σφαίρας, ἥτις ἐστὶ καὶ τοῦ ἡμικυκλίου διάμετρος δηλαδὴ καὶ τοῦ κύκλου, μείζων ἐστὶ πασῶν τῶν εἰς τὸν κύκλον ἢ τὴν σφαῖραν διαγομένων [εὐθειῶν].
Let then BCDE be the circle in the greater sphere, and FGH the circle in the lesser sphere; let two diameters in them, BD, CE, be drawn at right angles to one another; then, given the two circles BCDE, FGH about the same centre, let there be inscribed in the greater circle BCDE an equilateral polygon with an even number of sides which does not touch the lesser circle FGH, let BK, KL, LM ME be its sides in the quadrant BE. let KA be joined and carried through to N, let AO be set up from the point A at right angles to the plane of the circle BCDE, and let it meet the surface of the sphere at O, and through AO and each of the straight lines BD, KN let planes be carried; they will then make greatest circles on the surface of the sphere, for the reason stated.
ἔστω οὖν ἐν μὲν τῇ μείζονι σφαίρᾳ κύκλος ὁ ΒΓΔΕ, ἐν δὲ τῇ ἐλάσσονι σφαίρᾳ κύκλος ὁ ΖΗΘ, καὶ ἤχθωσαν αὐτῶν δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ ΒΔ, ΓΕ, καὶ δύο κύκλων περὶ τὸ αὐτὸ κέντρον ὄντων τῶν ΒΓΔΕ, ΖΗΘ εἰς τὸν μείζονα κύκλον τὸν ΒΓΔΕ πολύγωνον ἰσόπλευρον καὶ ἀρτιόπλευρον ἐγγεγράφθω μὴ ψαῦον τοῦ ἐλάσσονος κύκλου τοῦ ΖΗΘ, οὗ πλευραὶ ἔστωσαν ἐν τῷ ΒΕ τεταρτημορίῳ αἱ ΒΚ, ΚΛ, ΛΜ, ΜΕ, καὶ ἐπιζευχθεῖσα ἡ ΚΑ διήχθω ἐπὶ τὸ Ν, καὶ ἀνεστάτω ἀπὸ τοῦ Α σημείου τῷ τοῦ ΒΓΔΕ κύκλου ἐπιπέδῳ πρὸς ὀρθὰς ἡ ΑΞ καὶ συμβαλλέτω τῇ ἐπιφανείᾳ τῆς σφαίρας κατὰ τὸ Ξ, καὶ διὰ τῆς ΑΞ καὶ ἑκατέρας τῶν ΒΔ, ΚΝ ἐπίπεδα ἐκβεβλήσθω· ποιήσουσι δὴ διὰ τὰ εἰρημένα ἐπὶ τῆς ἐπιφανείας τῆς σφαίρας μεγίστους κύκλους.
Let them make such, and in them let BOD, KON be the semicircles on BD, KN.
ποιείτωσαν, ὧν ἡμικύκλια ἔστω ἐπὶ τῶν ΒΔ, ΚΝ διαμέτρων τὰ ΒΞΔ, ΚΞΝ.
Now, since OA is at right angles to the plane of the circle BCDE, therefore all the planes through OA are also at right angles to the plane of the circle BCDE; [] hence the semicircles BOD, KON are also at right angles to the plane of the circle BCDE.
καὶ ἐπεὶ ἡ ΞΑ ὀρθή ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον, καὶ πάντα ἄρα τὰ διὰ τῆς ΞΑ ἐπίπεδά ἐστιν ὀρθὰ πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον· ὥστε καὶ τὰ ΒΞΔ, ΚΞΝ ἡμικύκλια ὀρθά ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον.
And, since the semicircles BED, BOD, KON are equal, for they are on the equal diameters BD, KN, therefore the quadrants BE, BO, KO are also equal to one another.
καὶ ἐπεὶ ἴσα ἐστὶ τὰ ΒΕΔ, ΒΞΔ, ΚΞΝ ἡμικύκλια· ἐπὶ γὰρ ἴσων εἰσὶ διαμέτρων τῶν ΒΔ, ΚΝ· ἴσα ἐστὶ καὶ τὰ ΒΕ, ΒΞ, ΚΞ τεταρτημόρια ἀλλήλοις.
Therefore there are as many straight lines in the quadrants BO, KO equal to the straight lines BK, KL, LM, ME as there are sides of the polygon in the quadrant BE.
ὅσαι ἄρα εἰσὶν ἐν τῷ ΒΕ τεταρτημορίῳ πλευραὶ τοῦ πολυγώνου, τοσαῦταί εἰσι καὶ ἐν τοῖς ΒΞ, ΚΞ τεταρτημορίοις ἴσαι ταῖς ΒΚ, ΚΛ, ΛΜ, ΜΕ εὐθείαις.
Let them be inscribed, and let them be BP, PQ, QR, RO and KS, ST, TU, UO, let SP, TQ, UR be joined, and from P, S let perpendiculars be drawn to the plane of the circle BCDE; [] these will fall on BD, KN, the common sections of the planes, inasmuch as the planes of BOD, KON are also at right angles to the plane of the circle BCDE. [cf. ]
ἐγγεγράφθωσαν καὶ ἔστωσαν αἱ ΒΟ, ΟΠ, ΠΡ, ΡΞ, ΚΣ, ΣΤ, ΤΥ, ΥΞ, καὶ ἐπεζεύχθωσαν αἱ ΣΟ, ΤΠ, ΥΡ, καὶ ἀπὸ τῶν Ο, Σ ἐπὶ τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον κάθετοι ἤχθωσαν· πεσοῦνται δὴ ἐπὶ τὰς κοινὰς τομὰς τῶν ἐπιπέδων τὰς ΒΔ, ΚΝ, ἐπειδήπερ καὶ τὰ τῶν ΒΞΔ, ΚΞΝ ἐπίπεδα ὀρθά ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον.
Let them so fall, and let them be PV, SW, and let WV be joined.
πιπτέτωσαν, καὶ ἔστωσαν αἱ ΟΦ, ΣΧ, καὶ ἐπεζεύχθω ἡ ΧΦ.
Now since, in the equal semicircles BOD, KON, equal straight lines BP, KS have been cut off, and the perpendiculars PV, SW have been drawn, therefore PV is equal to SW, and BV to KW. [, ]
καὶ ἐπεὶ ἐν ἴσοις ἡμικυκλίοις τοῖς ΒΞΔ, ΚΞΝ ἴσαι ἀπειλημμέναι εἰσὶν αἱ ΒΟ, ΚΣ, καὶ κάθετοι ἠγμέναι εἰσὶν αἱ ΟΦ, ΣΧ, ἴση [ἄρα] ἐστὶν ἡ μὲν ΟΦ τῇ ΣΧ, ἡ δὲ ΒΦ τῇ ΚΧ.
But the whole BA is also equal to the whole KA; therefore the remainder VA is also equal to the remainder WA; therefore, as BV is to VA, so is KW to WA; therefore WV is parallel to KB. []
ἔστι δὲ καὶ ὅλη ἡ ΒΑ ὅλῃ τῇ ΚΑ ἴση· καὶ λοιπὴ ἄρα ἡ ΦΑ λοιπῇ τῇ ΧΑ ἐστιν ἴση· ἔστιν ἄρα ὡς ἡ ΒΦ πρὸς τὴν ΦΑ, οὕτως ἡ ΚΧ πρὸς τὴν ΧΑ· παράλληλος ἄρα ἐστὶν ἡ ΧΦ τῇ ΚΒ.
And, since each of the straight lines PV, SW is at right angles to the plane of the circle BCDE, therefore PV is parallel to SW. []
καὶ ἐπεὶ ἑκατέρα τῶν ΟΦ, ΣΧ ὀρθή ἐστι πρὸς τὸ τοῦ ΒΓΔΕ κύκλου ἐπίπεδον, παράλληλος ἄρα ἐστὶν ἡ ΟΦ τῇ ΣΧ. ἐδείχθη δὲ αὐτῇ καὶ ἴση·
But it was also proved equal to it; therefore WV, SP are also equal and parallel. []
καὶ αἱ ΧΦ, ΣΟ ἄρα ἴσαι εἰσὶ καὶ παράλληλοι.
And, since WV is parallel to SP, while WV is parallel to KB, therefore SP is also parallel to KB. []
καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΧΦ τῇ ΣΟ, ἀλλὰ ἡ ΧΦ τῇ ΚΒ ἐστι παράλληλος, καὶ ἡ ΣΟ ἄρα τῇ ΚΒ ἐστι παράλληλος.
And BP, KS join their extremities; therefore the quadrilateral KBPS is in one plane, inasmuch as, if two straight lines be parallel, and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallels. []
καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΒΟ, ΚΣ· τὸ ΚΒΟΣ ἄρα τετράπλευρον ἐν ἑνί ἐστιν ἐπιπέδῳ, ἐπειδήπερ, ἐὰν ὦσι δύο εὐθεῖαι παράλληλοι, καὶ ἐφʼ ἑκατέρας αὐτῶν ληφθῇ τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐν τῷ αὐτῷ ἐπιπέδῳ ἐστὶ ταῖς παραλλήλοις.
For the same reason each of the quadrilaterals SPQT, TQRU is also in one plane. But the triangle URO is also in one plane. []
διὰ τὰ αὐτὰ δὴ καὶ ἑκάτερον τῶν ΣΟΠΤ, ΤΠΡΥ τετραπλεύρων ἐν ἑνί ἐστιν ἐπιπέδῳ.
If then we conceive straight lines joined from the points P, S, Q, T, R, U to A, there will be constructed a certain polyhedral solid figure between the circumferences BO, KO, consisting of pyramids of which the quadrilaterals KBPS, SPQT, TQRU and the triangle URO are the bases and the point A the vertex.
ἔστι δὲ καὶ τὸ ΥΡΞ τρίγωνον ἐν ἑνὶ ἐπιπέδῳ. ἐὰν δὴ νοήσωμεν ἀπὸ τῶν Ο, Σ, Π, Τ, Ρ, Υ σημείων ἐπὶ τὸ Α ἐπιζευγνυμένας εὐθείας, συσταθήσεταί τι σχῆμα στερεὸν πολύεδρον μεταξὺ τῶν ΒΞ, ΚΞ περιφερειῶν ἐκ πυραμίδων συγκείμενον, ὧν βάσεις μὲν τὰ ΚΒΟΣ, ΣΟΠΤ, ΤΠΡΥ τετράπλευρα καὶ τὸ ΥΡΞ τρίγωνον, κορυφὴ δὲ τὸ Α σημεῖον.
And, if we make the same construction in the case of each of the sides KL, LM, ME as in the case of BK, and further in the case of the remaining three quadrants, there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids, of which the said quadrilaterals and the triangle URO, and the others corresponding to them, are the bases and the point A the vertex.
ἐὰν δὲ καὶ ἐπὶ ἑκάστης τῶν ΚΛ, ΛΜ, ΜΕ πλευρῶν καθάπερ ἐπὶ τῆς ΒΚ τὰ αὐτὰ κατασκευάσωμεν καὶ ἔτι ἐπὶ τῶν λοιπῶν τριῶν τεταρτημορίων, συσταθήσεταί τι σχῆμα πολύεδρον ἐγγεγραμμένον εἰς τὴν σφαῖραν πυραμίσι περιεχόμενον, ὧν βάσεις [μὲν] τὰ εἰρημένα τετράπλευρα καὶ τὸ ΥΡΞ τρίγωνον καὶ τὰ ὁμοταγῆ αὐτοῖς, κορυφὴ δὲ τὸ Α σημεῖον.
I say that the said polyhedron will not touch the lesser sphere at the surface on which the circle FGH is.
λέγω, ὅτι τὸ εἰρημένον πολύεδρον οὐκ ἐφάψεται τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν, ἐφʼ ἧς ἐστιν ὁ ΖΗΘ κύκλος.
Let AX be drawn from the point A perpendicular to the plane of the quadrilateral KBPS, and let it meet the plane at the point X; [] let XB, XK be joined.
ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὸ τοῦ ΚΒΟΣ τετραπλεύρου ἐπίπεδον κάθετος ἡ ΑΨ καὶ συμβαλλέτω τῷ ἐπιπέδῳ κατὰ τὸ Ψ σημεῖον, καὶ ἐπεζεύχθωσαν αἱ ΨΒ, ΨΚ.
Then, since AX is at right angles to the plane of the quadrilateral KBPS, therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. []
καὶ ἐπεὶ ἡ ΑΨ ὀρθή ἐστι πρὸς τὸ τοῦ ΚΒΟΣ τετραπλεύρου ἐπίπεδον, καὶ πρὸς πάσας ἄρα τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ τοῦ τετραπλεύρου ἐπιπέδῳ ὀρθή ἐστιν.
Therefore AX is at right angles to each of the straight lines BX, XK.
ἡ ΑΨ ἄρα ὀρθή ἐστι πρὸς ἑκατέραν τῶν ΒΨ, ΨΚ.
And, since AB is equal to AK, the square on AB is also equal to the square on AK.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΑΚ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΑΒ τῷ ἀπὸ τῆς ΑΚ.
And the squares on AX, XB are equal to the square on AB, for the angle at X is right; [] and the squares on AX, XK are equal to the square on AK. [id.]
καί ἐστι τῷ μὲν ἀπὸ τῆς ΑΒ ἴσα τὰ ἀπὸ τῶν ΑΨ, ΨΒ· ὀρθὴ γὰρ ἡ πρὸς τῷ Ψ· τῷ δὲ ἀπὸ τῆς ΑΚ ἴσα τὰ ἀπὸ τῶν ΑΨ, ΨΚ.
Therefore the squares on AX, XB are equal to the squares on AX, XK.
τὰ ἄρα ἀπὸ τῶν ΑΨ, ΨΒ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΑΨ, ΨΚ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΑΨ·
Let the square on AX be subtracted from each; therefore the remainder, the square on BX, is equal to the remainder, the square on XK; therefore BX is equal to XK. Similarly we can prove that the straight lines joined from X to P, S are equal to each of the straight lines BX, XK.
λοιπὸν ἄρα τὸ ἀπὸ τῆς ΒΨ λοιπῷ τῷ ἀπὸ τῆς ΨΚ ἴσον ἐστίν· ἴση ἄρα ἡ ΒΨ τῇ ΨΚ. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ἀπὸ τοῦ Ψ ἐπὶ τὰ Ο, Σ ἐπιζευγνύμεναι εὐθεῖαι ἴσαι εἰσὶν ἑκατέρᾳ τῶν ΒΨ, ΨΚ.
Therefore the circle described with centre X and distance one of the straight lines XB, XK will pass through P, S also, and KBPS will be a quadrilateral in a circle.
ὁ ἄρα κέντρῳ τῷ Ψ καὶ διαστήματι ἑνὶ τῶν ΨΒ, ΨΚ γραφόμενος κύκλος ἥξει καὶ διὰ τῶν Ο, Σ, καὶ ἔσται ἐν κύκλῳ τὸ ΚΒΟΣ τετράπλευρον. καὶ ἐπεὶ μείζων ἐστὶν ἡ ΚΒ τῆς ΧΦ, ἴση δὲ ἡ ΧΦ τῇ ΣΟ, μείζων ἄρα ἡ ΚΒ τῆς ΣΟ.
Now, since KB is greater than WV, while WV is equal to SP, therefore KB is greater than SP. But KB is equal to each of the straight lines KS, BP; therefore each of the straight lines KS, BP is greater than SP.
ἴση δὲ ἡ ΚΒ ἑκατέρᾳ τῶν ΚΣ, ΒΟ· καὶ ἑκατέρα ἄρα τῶν ΚΣ, ΒΟ τῆς ΣΟ μείζων ἐστίν.
And, since KBPS is a quadrilateral in a circle, and KB, BP, KS are equal, and PS less, and BX is the radius of the circle, therefore the square on KB is greater than double of the square on BX. Let KZ be drawn from K perpendicular to BV.
καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΚΒΟΣ, καὶ ἴσαι αἱ ΚΒ, ΒΟ, ΚΣ, καὶ ἐλάττων ἡ ΟΣ, καὶ ἐκ τοῦ κέντρου τοῦ κύκλου ἐστὶν ἡ ΒΨ, τὸ ἄρα ἀπὸ τῆς ΚΒ τοῦ ἀπὸ τῆς ΒΨ μεῖζόν ἐστιν ἢ διπλάσιον. ἤχθω ἀπὸ τοῦ Κ ἐπὶ τὴν ΒΦ κάθετος ἡ ΚΩ.
Then, since BD is less than double of DZ, and, as BD is to DZ, so is the rectangle DB, BZ to the rectangle DZ, ZB, if a square be described upon BZ and the parallelogram on ZD be completed, then the rectangle DB, BZ is also less than double of the rectangle DZ, ZB.
καὶ ἐπεὶ ἡ ΒΔ τῆς ΔΩ ἐλάττων ἐστὶν ἢ διπλῆ, καί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΩ, οὕτως τὸ ὑπὸ τῶν ΔΒ, ΒΩ πρὸς τὸ ὑπὸ [τῶν] ΔΩ, ΩΒ, ἀναγραφομένου ἀπὸ τῆς ΒΩ τετραγώνου καὶ συμπληρουμένου τοῦ ἐπὶ τῆς ΩΔ παραλληλογράμμου καὶ τὸ ὑπὸ ΔΒ, ΒΩ ἄρα τοῦ ὑπὸ ΔΩ, ΩΒ ἔλαττόν ἐστιν ἢ διπλάσιον.
And, if KD be joined, the rectangle DB, BZ is equal to the square on BK, and the rectangle DZ, ZB equal to the square on KZ; [, ] therefore the square on KB is less than double of the square on KZ.
καί ἐστι τῆς ΚΔ ἐπιζευγνυμένης τὸ μὲν ὑπὸ ΔΒ, ΒΩ ἴσον τῷ ἀπὸ τῆς ΒΚ, τὸ δὲ ὑπὸ τῶν ΔΩ, ΩΒ ἴσον τῷ ἀπὸ τῆς ΚΩ· τὸ ἄρα ἀπὸ τῆς ΚΒ τοῦ ἀπὸ τῆς ΚΩ ἔλασσόν ἐστιν ἢ διπλάσιον. ἀλλὰ τὸ ἀπὸ τῆς ΚΒ τοῦ ἀπὸ τῆς ΒΨ μεῖζόν ἐστιν ἢ διπλάσιον·
But the square on KB is greater than double of the square on BX; therefore the square on KZ is greater than the square on BX. And, since BA is equal to KA, the square on BA is equal to the square on AK.
μεῖζον ἄρα τὸ ἀπὸ τῆς ΚΩ τοῦ ἀπὸ τῆς ΒΨ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΑ τῇ ΚΑ, ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΑ τῷ ἀπὸ τῆς ΑΚ.
And the squares on BX, XA are equal to the square on BA, and the squares on KZ, ZA equal to the square on KA; [] therefore the squares on BX, XA are equal to the squares on KZ, ZA, and of these the square on KZ is greater than the square on BX; therefore the remainder, the square on ZA, is less than the square on XA.
καί ἐστι τῷ μὲν ἀπὸ τῆς ΒΑ ἴσα τὰ ἀπὸ τῶν ΒΨ, ΨΑ, τῷ δὲ ἀπὸ τῆς ΚΑ ἴσα τὰ ἀπὸ τῶν ΚΩ, ΩΑ· τὰ ἄρα ἀπὸ τῶν ΒΨ, ΨΑ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΚΩ, ΩΑ, ὧν τὸ ἀπὸ τῆς ΚΩ μεῖζον τοῦ ἀπὸ τῆς ΒΨ· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΩΑ ἔλασσόν ἐστι τοῦ ἀπὸ τῆς ΨΑ. μείζων ἄρα ἡ ΑΨ τῆς ΑΩ· πολλῷ ἄρα ἡ ΑΨ μείζων ἐστὶ τῆς ΑΗ.
Therefore AX is greater than AZ; therefore AX is much greater than AG.
καί ἐστιν ἡ μὲν ΑΨ ἐπὶ μίαν τοῦ πολυέδρου βάσιν, ἡ δὲ ΑΗ ἐπὶ τὴν τῆς ἐλάσσονος σφαίρας ἐπιφάνειαν·
And AX is the perpendicular on one base of the polyhedron, and AG on the surface of the lesser sphere; hence the polyhedron will not touch the lesser sphere on its surface.
ὥστε τὸ πολύεδρον οὐ ψαύσει τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν.
Therefore, given two spheres about the same centre, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface. Q. E. F.
δύο ἄρα σφαιρῶν περὶ τὸ αὐτὸ κέντρον οὐσῶν εἰς τὴν μείζονα σφαῖραν στερεὸν πολύεδρον ἐγγέγραπται μὴ ψαῦον τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν· ὅπερ ἔδει ποιῆσαι.
Porism. But if in another sphere also a polyhedral solid be inscribed similar to the solid in the sphere BCDE, the polyhedral solid in the sphere BCDE has to the polyhedral solid in the other sphere the ratio triplicate of that which the diameter of the sphere BCDE has to the diameter of the other sphere.
Πόρισμα ἐὰν δὲ καὶ εἰς ἑτέραν σφαῖραν τῷ ἐν τῇ ΒΓΔΕ σφαίρᾳ στερεῷ πολυέδρῳ ὅμοιον στερεὸν πολύεδρον ἐγγραφῇ, τὸ ἐν τῇ ΒΓΔΕ σφαίρᾳ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ ἑτέρᾳ σφαίρᾳ στερεὸν πολύεδρον τριπλασίονα λόγον ἔχει, ἤπερ ἡ τῆς ΒΓΔΕ σφαίρας διάμετρος πρὸς τὴν τῆς ἑτέρας σφαίρας διάμετρον.
For, the solids being divided into their pyramids similar in multitude and arrangement, the pyramids will be similar.
διαιρεθέντων γὰρ τῶν στερεῶν εἰς τὰς ὁμοιοπληθεῖς καὶ ὁμοιοταγεῖς πυραμίδας ἔσονται αἱ πυραμίδες ὅμοιαι.
But similar pyramids are to one another in the triplicate ratio of their corresponding sides; [] therefore the pyramid of which the quadrilateral KBPS is the base, and the point A the vertex, has to the similarly arranged pyramid in the other sphere the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of that which the radius AB of the sphere about A as centre has to the radius of the other sphere.
αἱ δὲ ὅμοιαι πυραμίδες πρὸς ἀλλήλας ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν· ἡ ἄρα πυραμίς, ἧς βάσις μέν ἐστι τὸ ΚΒΟΣ τετράπλευρον, κορυφὴ δὲ τὸ Α σημεῖον, πρὸς τὴν ἐν τῇ ἑτέρᾳ σφαίρᾳ ὁμοιοταγῆ πυραμίδα τριπλασίονα λόγον ἔχει, ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἤπερ ἡ ΑΒ ἐκ τοῦ κέντρου τῆς σφαίρας τῆς περὶ κέντρον τὸ Α πρὸς τὴν ἐκ τοῦ κέντρου τῆς ἑτέρας σφαίρας.
Similarly also each pyramid of those in the sphere about A as centre has to each similarly arranged pyramid of those in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere.
ὁμοίως καὶ ἑκάστη πυραμὶς τῶν ἐν τῇ περὶ κέντρον τὸ Α σφαίρᾳ πρὸς ἑκάστην ὁμοταγῆ πυραμίδα τῶν ἐν τῇ ἑτέρᾳ σφαίρᾳ τριπλασίονα λόγον ἕξει, ἤπερ ἡ ΑΒ πρὸς τὴν ἐκ τοῦ κέντρου τῆς ἑτέρας σφαίρας.
And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [] hence the whole polyhedral solid in the sphere about A as centre has to the whole polyhedral solid in the other sphere the ratio triplicate of that which AB has to the radius of the other sphere, that is, of that which the diameter BD has to the diameter of the other sphere. Q. E. D.
καὶ ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ὥστε ὅλον τὸ ἐν τῇ περὶ κέντρον τὸ Α σφαίρᾳ στερεὸν πολύεδρον πρὸς ὅλον τὸ ἐν τῇ ἑτέρᾳ [σφαίρᾳ] στερεὸν πολύεδρον τριπλασίονα λόγον ἕξει, ἤπερ ἡ ΑΒ πρὸς τὴν ἐκ τοῦ κέντρου τῆς ἑτέρας σφαίρας, τουτέστιν ἤπερ ἡ ΒΔ διάμετρος πρὸς τὴν τῆς ἑτέρας σφαίρας διάμετρον· ὅπερ ἔδει δεῖξαι.
18. Spheres are to one another in the triplicate ratio of their respective diameters. Let the spheres ABC, DEF be conceived, and let BC, EF be their diameters; I say that the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF. For, if the sphere ABC has not to the sphere DEF the ratio triplicate of that which BC has to EF, then the sphere ABC will have either to some less sphere than the sphere DEF, or to a greater, the ratio triplicate of that which BC has to EF. First, let it have that ratio to a less sphere GHK, let DEF be conceived about the same centre with GHK, let there be inscribed in the greater sphere DEF a polyhedral solid which does not touch the lesser sphere GHK at its surface, [] and let there also be inscribed in the sphere ABC a polyhedral solid similar to the polyhedral solid in the sphere DEF; therefore the polyhedral solid in ABC has to the polyhedral solid in DEF the ratio triplicate of that which BC has to EF. [] But the sphere ABC also has to the sphere GHK the ratio triplicate of that which BC has to EF; therefore, as the sphere ABC is to the sphere GHK, so is the polyhedral solid in the sphere ABC to the polyhedral solid in the sphere DEF; and, alternately, as the sphere ABC is to the polyhedron in it, so is the sphere GHK to the polyhedral solid in the sphere DEF. [] But the sphere ABC is greater than the polyhedron in it; therefore the sphere GHK is also greater than the polyhedron in the sphere DEF. But it is also less, for it is enclosed by it. Therefore the sphere ABC has not to a less sphere than the sphere DEF the ratio triplicate of that which the diameter BC has to EF. Similarly we can prove that neither has the sphere DEF to a less sphere than the sphere ABC the ratio triplicate of that which EF has to BC. I say next that neither has the sphere ABC to any greater sphere than the sphere DEF the ratio triplicate of that which BC has to EF. For, if possible, let it have that ratio to a greater, LMN; therefore, inversely, the sphere LMN has to the sphere ABC the ratio triplicate of that which the diameter EF has to the diameter BC. But, inasmuch as LMN is greater than DEF, therefore, as the sphere LMN is to the sphere ABC, so is the sphere DEF to some less sphere than the sphere ABC, as was before proved. [] Therefore the sphere DEF also has to some less sphere than the sphere ABC the ratio triplicate of that which EF has to BC: which was proved impossible. Therefore the sphere ABC has not to any sphere greater than the sphere DEF the ratio triplicate of that which BC has to EF. But it was proved that neither has it that ratio to a less sphere. Therefore the sphere ABC has to the sphere DEF the ratio triplicate of that which BC has to EF. Q. E. D. I have already given, in the note to, the evidence upon which the construction of the five regular solids is attributed to the Pythagoreans. Some of them, the cube, the tetrahedron (which is nothing but a pyramid), and the octahedron (which is only a double pyramid with a square base), cannot but have been known to the Egyptians. And it appears that dodecahedra have been found, of bronze or other material, which may belong to periods earlier than Pythagoras' time by some centuries (for references see Cantor's Geschichte der Mathematik I_{3}, pp. 175-6). It is true that the author of the scholium No. I to Eucl. XIII. says that the Book is aboutthe five so-called Platonic figures, which however do not belong to Plato, three of the aforesaid five figures being due to the Pythagoreans, namely the cube, the pyramid and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus.This statement (taken probably from Geminus) may perhaps rest on the fact that Theaetetus was the first to write at any length about the two last-mentioned solids. We are told indeed by Suidas (S. V. Θεαίτητος) that Theaetetusfirst wrote on the ’five solids’ as they are called.This no doubt means that Theaetetus was the first to write a complete and systematic treatise on all the regular solids; it does not exclude the possibility that Hippasus or others had already written on the dodecahedron. The fact that Theaetetus wrote upon the regular solids agrees very well with the evidence which we possess of his contributions to the theory of irrationals, the connexion between which and the investigation of the regular solids is seen in Euclid's Book XIII. Theaetetus flourished about 380 B.C., and his work on the regular solids was soon followed by another, that of Aristaeus, an elder contemporary of Euclid, who also wrote an important book on Solid Loci, i.e. on conics treated as loci. This Aristaeus (known asthe elder) wrote in the period about 320 B.C. We hear of his Comparison of the five regular solids from Hypsicles (2nd cent. B.C.), the writer of the short book commonly included in the editions of the Elements as Book XIV. Hypsicles gives in this Book some six propositions supplementing Eucl. XIII.; and he introduces the second of the propositions (Heiberg's Euclid, Vol. v. p. 6) as follows: The same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron when both are inscribed in the same sphere. This is proved by Aristaeus in the book entitled Comparison of the five figures. Hypsicles proceeds (pp. 7 sqq.) to give a proof of this theorem. Allman pointed out (Greek Geometry from Thales to Euclid, 1889, pp. 201-2) that this proof depends on eight theorems, six of which appear in Euclid's Book XIII. (in Propositions,,, ); two other propositions not mentioned by Allman are also used, namely. This seems, as Allman says, to confirm the inference of Bretschneider (p. 171) that, as Aristaeus' work was the newest and latest in which, before Euclid's time, this subject was treated, we have in Eucl. XIII. at least a partial recapitulation of the contents of the treatise of Aristaeus. After Euclid, Apollonius wrote on the comparison of the dodecahedron and the icosahedron inscribed in one and the same sphere. This we also learn from Hypsicles, who says in the next words following those about Aristaeus above quoted:But it is proved by Apollonius in the second edition of his Comparison of the dodecahedron with the icosahedron that, as the surface of the dodecahedron is to the surface of the icosahedron [inscribed in the same sphere], so is the dodecahedron itself [i.e. its volume] to the icosahedron, because the perpendicular is the same from the centre of the sphere to the pentagon of the dodecahedron and to the triangle of the icosahedron. If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square on the half. For let the straight line AB be cut in extreme and mean ratio at the point C, and let AC be the greater segment; let the straight line AD be produced in a straight line with CA, and let AD be made half of AB; I say that the square on CD is five times the square on AD. For let the squares AE, DF be described on AB, DC, and let the figure in DF be drawn; let FC be carried through to G. Now, since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [, ] And CE is the rectangle AB, BC, and FH the square on AC; therefore CE is equal to FH. And, since BA is double of AD, while BA is equal to KA, and AD to AH, therefore KA is also double of AH. But, as KA is to AH, so is CK to CH; [] therefore CK is double of CH. But LH, HC are also double of CH. Therefore KC is equal to LH, HC. But CE was also proved equal to HF; therefore the whole square AE is equal to the gnomon MNO. And, since BA is double of AD, the square on BA is quadruple of the square on AD, that is, AE is quadruple of DH. But AE is equal to the gnomon MNO; therefore the gnomon MNO is also quadruple of AP; therefore the whole DF is five times AP. And DF is the square on DC, and AP the square on DA; therefore the square on CD is five times the square on DA. Therefore etc. Q. E. D.
18. αἱ σφαῖραι πρὸς ἀλλήλας ἐν τριπλασίονι λόγῳ εἰσὶ τῶν ἰδίων διαμέτρων. Νενοήσθωσαν σφαῖραι αἱ ΑΒΓ, ΔΕΖ, διάμετροι δὲ αὐτῶν αἱ ΒΓ, ΕΖ· λέγω, ὅτι ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. εἰ γὰρ μὴ ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ, ἕξει ἄρα ἡ ΑΒΓ σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἢ πρὸς μείζονα ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἐχέτω πρότερον πρὸς ἐλάσσονα τὴν ΗΘΚ, καὶ νενοήσθω ἡ ΔΕΖ τῇ ΗΘΚ περὶ τὸ αὐτὸ κέντρον, καὶ ἐγγεγράφθω εἰς τὴν μείζονα σφαῖραν τὴν ΔΕΖ στερεὸν πολύεδρον μὴ ψαῦον τῆς ἐλάσσονος σφαίρας τῆς ΗΘΚ κατὰ τὴν ἐπιφάνειαν, ἐγγεγράφθω δὲ καὶ εἰς τὴν ΑΒΓ σφαῖραν τῷ ἐν τῇ ΔΕΖ σφαίρᾳ στερεῷ πολυέδρῳ ὅμοιον στερεὸν πολύεδρον· τὸ ἄρα ἐν τῇ ΑΒΓ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ ΔΕΖ στερεὸν πολύεδρον τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἔχει δὲ καὶ ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΗΘΚ σφαῖραν τριπλασίονα λόγον ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ· ἔστιν ἄρα ὡς ἡ ΑΒΓ σφαῖρα πρὸς τὴν ΗΘΚ σφαῖραν, οὕτως τὸ ἐν τῇ ΑΒΓ σφαίρᾳ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ ΔΕΖ σφαίρᾳ στερεὸν πολύεδρον· ἐναλλὰξ [ἄρα] ὡς ἡ ΑΒΓ σφαῖρα πρὸς τὸ ἐν αὐτῇ πολύεδρον, οὕτως ἡ ΗΘΚ σφαῖρα πρὸς τὸ ἐν τῇ ΔΕΖ σφαίρᾳ στερεὸν πολύεδρον. μείζων δὲ ἡ ΑΒΓ σφαῖρα τοῦ ἐν αὐτῇ πολυέδρου· μείζων ἄρα καὶ ἡ ΗΘΚ σφαῖρα τοῦ ἐν τῇ ΔΕΖ σφαίρᾳ πολυέδρου. ἀλλὰ καὶ ἐλάττων· ἐμπεριέχεται γὰρ ὑπʼ αὐτοῦ. οὐκ ἄρα ἡ ΑΒΓ σφαῖρα πρὸς ἐλάσσονα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ διάμετρος πρὸς τὴν ΕΖ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἡ ΔΕΖ σφαῖρα πρὸς ἐλάσσονα τῆς ΑΒΓ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΒΓ. λέγω δή, ὅτι οὐδὲ ἡ ΑΒΓ σφαῖρα πρὸς μείζονά τινα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. εἰ γὰρ δυνατόν, ἐχέτω πρὸς μείζονα τὴν ΛΜΝ· ἀνάπαλιν ἄρα ἡ ΛΜΝ σφαῖρα πρὸς τὴν ΑΒΓ σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ διάμετρος πρὸς τὴν ΒΓ διάμετρον. ὡς δὲ ἡ ΛΜΝ σφαῖρα πρὸς τὴν ΑΒΓ σφαῖραν, οὕτως ἡ ΔΕΖ σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΑΒΓ σφαίρας, ἐπειδήπερ μείζων ἐστὶν ἡ ΛΜΝ τῆς ΔΕΖ, ὡς ἔμπροσθεν ἐδείχθη. καὶ ἡ ΔΕΖ ἄρα σφαῖρα πρὸς ἐλάσσονά τινα τῆς ΑΒΓ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΒΓ· ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα ἡ ΑΒΓ σφαῖρα πρὸς μείζονά τινα τῆς ΔΕΖ σφαίρας τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἐδείχθη δέ, ὅτι οὐδὲ πρὸς ἐλάσσονα. ἡ ἄρα ΑΒΓ σφαῖρα πρὸς τὴν ΔΕΖ σφαῖραν τριπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ· ὅπερ ἔδει δεῖξαι. ἐὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ μεῖζον τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου. εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τετμήσθω κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα τὸ ΑΓ, καὶ ἐκβεβλήσθω ἐπʼ εὐθείας τῇ ΓΑ εὐθεῖα ἡ ΑΔ, καὶ κείσθω τῆς ΑΒ ἡμίσεια ἡ ΑΔ· λέγω, ὅτι πενταπλάσιόν ἐστι τὸ ἀπὸ τῆς ΓΔ τοῦ ἀπὸ τῆς ΔΑ. ἀναγεγράφθωσαν γὰρ ἀπὸ τῶν ΑΒ, ΔΓ τετράγωνα τὰ ΑΕ, ΔΖ, καὶ καταγεγράφθω ἐν τῷ ΔΖ τὸ σχῆμα, καὶ διήχθω ἡ ΖΓ ἐπὶ τὸ Η. καὶ ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Γ, τὸ ἄρα ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΑΒΓ τὸ ΓΕ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΖΘ· ἴσον ἄρα τὸ ΓΕ τῷ ΖΘ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΒΑ τῆς ΑΔ, ἴση δὲ ἡ μὲν ΒΑ τῇ ΚΑ, ἡ δὲ ΑΔ τῇ ΑΘ, διπλῆ ἄρα καὶ ἡ ΚΑ τῆς ΑΘ. ὡς δὲ ἡ ΚΑ πρὸς τὴν ΑΘ, οὕτως τὸ ΓΚ πρὸς τὸ ΓΘ· διπλάσιον ἄρα τὸ ΓΚ τοῦ ΓΘ. εἰσὶ δὲ καὶ τὰ ΛΘ, ΘΓ διπλάσια τοῦ ΓΘ. ἴσον ἄρα τὸ ΚΓ τοῖς ΛΘ, ΘΓ. ἐδείχθη δὲ καὶ τὸ ΓΕ τῷ ΘΖ ἴσον· ὅλον ἄρα τὸ ΑΕ τετράγωνον ἴσον ἐστὶ τῷ ΜΝΞ γνώμονι. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΒΑ τῆς ΑΔ, τετραπλάσιόν ἐστι τὸ ἀπὸ τῆς ΒΑ τοῦ ἀπὸ τῆς ΑΔ, τουτέστι τὸ ΑΕ τοῦ ΔΘ. ἴσον δὲ τὸ ΑΕ τῷ ΜΝΞ γνώμονι· καὶ ὁ ΜΝΞ ἄρα γνώμων τετραπλάσιός ἐστι τοῦ ΑΟ· ὅλον ἄρα τὸ ΔΖ πενταπλάσιόν ἐστι τοῦ ΑΟ. καί ἐστι τὸ μὲν ΔΖ τὸ ἀπὸ τῆς ΔΓ, τὸ δὲ ΑΟ τὸ ἀπὸ τῆς ΔΑ· τὸ ἄρα ἀπὸ τῆς ΓΔ πενταπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΔΑ. ἐὰν ἄρα εὐθεῖα ἄκρον καὶ μέσον λόγον τμηθῇ, τὸ μεῖζον τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον δύναται τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου· ὅπερ ἔδει δεῖξαι.