§13.prop.17#1δωδεκάεδρον συστήσασθαι καὶ σφαίρᾳ περιλαβεῖν, ᾗ καὶ τὰ προειρημένα σχήματα, καὶ δεῖξαι, ὅτι ἡ τοῦ δωδεκαέδρου πλευρὰ ἄλογός ἐστιν ἡ καλουμένη ἀποτομή.
To construct a dodecahedron and comprehend it in a sphere, like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome.
Ἐκκείσθωσαν τοῦ προειρημένου κύβου δύο ἐπίπεδα πρὸς ὀρθὰς ἀλλήλοις τὰ ΑΒΓΔ, ΓΒΕΖ, καὶ τετμήσθω ἑκάστη τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ, ΕΖ, ΕΒ, ΖΓ πλευρῶν δίχα κατὰ τὰ Η, Θ, Κ, Λ, Μ, Ν, Ξ, καὶ ἐπεζεύχθωσαν αἱ ΗΚ, ΘΛ, ΜΘ, ΝΞ, καὶ τετμήσθω ἑκάστη τῶν ΝΟ, ΟΞ, ΘΠ ἄκρον καὶ μέσον λόγον κατὰ τὰ Ρ, Σ, Τ σημεῖα, καὶ ἔστω αὐτῶν μείζονα τμήματα τὰ ΡΟ, ΟΣ, ΤΠ, καὶ ἀνεστάτωσαν ἀπὸ τῶν Ρ, Σ, Τ σημείων τοῖς τοῦ κύβου ἐπιπέδοις πρὸς ὀρθὰς ἐπὶ τὰ ἐκτὸς μέρη τοῦ κύβου αἱ ΡΥ, ΣΦ, ΤΧ, καὶ κείσθωσαν ἴσαι ταῖς ΡΟ, ΟΣ, ΤΠ, καὶ ἐπεζεύχθωσαν αἱ ΥΒ, ΒΧ, ΧΓ, ΓΦ, ΦΥ. λέγω, ὅτι τὸ ΥΒΧΓΦ πεντάγωνον ἰσόπλευρόν τε καὶ ἐν ἑνὶ ἐπιπέδῳ καὶ ἔτι ἰσογώνιόν ἐστιν.
Let two planes of the aforesaid cube, at right angles to one another, be laid down, namely ABGD, GBEZ, and let each of the sides AB, BG, GD, DA, EZ, EB, ZG be bisected at the points H, Th, K, L, M, N, Xi, and let HK, ThL, MTh, NXi be joined, and let each of the straight lines NO, OXi, ThP be cut in extreme and mean ratio at the points R, S, T, and let their greater segments be RO, OS, TP, and let RY, SPhi, TX be set up from the points R, S, T at right angles to the planes of the cube towards the outer parts of the cube, and let them be made equal to RO, OS, TP, and let YB, BX, XG, GPhi, PhiY be joined. I say that the pentagon YBXGPhi is equilateral, in one plane, and further equiangular.
ἐπεζεύχθωσαν γὰρ αἱ ΡΒ, ΣΒ, ΦΒ. καὶ ἐπεὶ εὐθεῖα ἡ ΝΟ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ρ, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΡΟ, τὰ ἄρα ἀπὸ τῶν ΟΝ, ΝΡ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΡΟ. ἴση δὲ ἡ μὲν ΟΝ τῇ ΝΒ, ἡ δὲ ΟΡ τῇ ΡΥ· τὰ ἄρα ἀπὸ τῶν ΒΝ, ΝΡ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΡΥ. τοῖς δὲ ἀπὸ τῶν ΒΝ, ΝΡ τὸ ἀπὸ τῆς ΒΡ ἐστιν ἴσον·
For let RB, SB, PhiB be joined. And since the straight line NO is cut in extreme and mean ratio at R, and its greater segment is RO, therefore the squares on ON, NR are triple of the square on RO. And ON is equal to NB, and OR to RY; therefore the squares on BN, NR are triple of the square on RY.
τὸ ἄρα ἀπὸ τῆς ΒΡ τριπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΡΥ· ὥστε τὰ ἀπὸ τῶν ΒΡ, ΡΥ τετραπλάσιά ἐστι τοῦ ἀπὸ τῆς ΡΥ. τοῖς δὲ ἀπὸ τῶν ΒΡ, ΡΥ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΒΥ·
But the square on BR is equal to the squares on BN, NR; therefore the square on BR is triple of the square on RY; so that the squares on BR, RY are quadruple of the square on RY.
τὸ ἄρα ἀπὸ τῆς ΒΥ τετραπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΥΡ·
And the square on BY is equal to the squares on BR, RY; therefore the square on BY is quadruple of the square on RY; therefore BY is double of RY.
διπλῆ ἄρα ἐστὶν ἡ ΒΥ τῆς ΡΥ. ἔστι δὲ καὶ ἡ ΦΥ τῆς ΥΡ διπλῆ, ἐπειδήπερ καὶ ἡ ΣΡ τῆς ΟΡ, τουτέστι τῆς ΡΥ, ἐστι διπλῆ·
And PhiY is also double of RY, since SR is also double of OR, that is, of RY; therefore BY is equal to YPhi.
ἴση ἄρα ἡ ΒΥ τῇ ΥΦ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΒΧ, ΧΓ, ΓΦ ἑκατέρᾳ τῶν ΒΥ, ΥΦ ἐστιν ἴση.
Similarly indeed it will be proved that each of the BX, XG, GPhi is also equal to each of BY, YPhi.
ἰσόπλευρον ἄρα ἐστὶ τὸ ΒΥΦΓΧ πεντάγωνον.
Therefore the pentagon BYPhiGX is equilateral.
λέγω δή, ὅτι καὶ ἐν ἑνί ἐστιν ἐπιπέδῳ.
I say then, that it is also in one plane.
ἤχθω γὰρ ἀπὸ τοῦ Ο ἑκατέρᾳ τῶν ΡΥ, ΣΦ παράλληλος ἐπὶ τὰ ἐκτὸς τοῦ κύβου μέρη ἡ ΟΨ, καὶ ἐπεζεύχθωσαν αἱ ΨΘ, ΘΧ· λέγω, ὅτι ἡ ΨΘΧ εὐθεῖά ἐστιν.
For let OPsi be drawn from O parallel to each of RY, SPhi towards the outer parts of the cube, and let PsiTh, ThX be joined; I say that PsiThX is a straight line.
ἐπεὶ γὰρ ἡ ΘΠ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Τ, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστιν ἡ ΠΤ, ἔστιν ἄρα ὡς ἡ ΘΠ πρὸς τὴν ΠΤ, οὕτως ἡ ΠΤ πρὸς τὴν ΤΘ. ἴση δὲ ἡ μὲν ΘΠ τῇ ΘΟ, ἡ δὲ ΠΤ ἑκατέρᾳ τῶν ΤΧ, ΟΨ· ἔστιν ἄρα ὡς ἡ ΘΟ πρὸς τὴν ΟΨ, οὕτως ἡ ΧΤ πρὸς τὴν ΤΘ. καί ἐστι παράλληλος ἡ μὲν ΘΟ τῇ ΤΧ·
For since ThP is cut in extreme and mean ratio at T, and its greater segment is PT, therefore, as ThP is to PT, so is PT to TTh. And ThP is equal to ThO, and PT is equal to each of TX, OPsi; therefore, as ThO is to OPsi, so is XT to TTh.
ἑκατέρα γὰρ αὐτῶν τῷ ΒΔ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν· ἡ δὲ ΤΘ τῇ ΟΨ· ἑκατέρα γὰρ αὐτῶν τῷ ΒΖ ἐπιπέδῳ πρὸς ὀρθάς ἐστιν.
And ThO is parallel to TX; for each of them is at right angles to the plane BD; and TTh is parallel to OPsi; for each of them is at right angles to the plane BZ.
ἐὰν δὲ δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν, ὡς τὰ ΨΟΘ, ΘΤΧ, τὰς δύο πλευρὰς ταῖς δυσὶν ἀνάλογον ἔχοντα, ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ εὐθεῖαι ἐπʼ εὐθείας ἔσονται· ἐπʼ εὐθείας ἄρα ἐστὶν ἡ ΨΘ τῇ ΘΧ. πᾶσα δὲ εὐθεῖα ἐν ἑνί ἐστιν ἐπιπέδῳ· ἐν ἑνὶ ἄρα ἐπιπέδῳ ἐστὶ τὸ ΥΒΧΓΦ πεντάγωνον.
And if two triangles, like PsiOTh, ThTX, are joined at one angle, having two sides proportional to two sides, so that their homologous sides are also parallel, the remaining straight lines will be in a straight line; therefore PsiTh is in a straight line with ThX. And every straight line is in one plane; therefore the pentagon YBXGPhi is in one plane.
λέγω δή, ὅτι καὶ ἰσογώνιόν ἐστιν.
I say then, that it is also equiangular.
ἐπεὶ γὰρ εὐθεῖα γραμμὴ ἡ ΝΟ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ρ, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΟΡ, ἴση δὲ ἡ ΟΡ τῇ ΟΣ, ἡ ΝΣ ἄρα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ο, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΝΟ·
For since the straight line NO is cut in extreme and mean ratio at R, and its greater segment is OR, and OR is equal to OS, therefore NS is cut in extreme and mean ratio at O, and its greater segment is NO; therefore the squares on NS, SO are triple of the square on NO.
τὰ ἄρα ἀπὸ τῶν ΝΣ, ΣΟ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΟ. ἴση δὲ ἡ μὲν ΝΟ τῇ ΝΒ, ἡ δὲ ΟΣ τῇ ΣΦ· τὰ ἄρα ἀπὸ τῶν ΝΣ, ΣΦ τετράγωνα τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΒ· ὥστε τὰ ἀπὸ τῶν ΦΣ, ΣΝ, ΝΒ τετραπλάσιά ἐστι τοῦ ἀπὸ τῆς ΝΒ. τοῖς δὲ ἀπὸ τῶν ΣΝ, ΝΒ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΣΒ·
And NO is equal to NB, and OS to SPhi; therefore the squares on NS, SPhi are triple of the square on NB; so that the squares on PhiS, SN, NB are quadruple of the square on NB.
τὰ ἄρα ἀπὸ τῶν ΒΣ, ΣΦ, τουτέστι τὸ ἀπὸ τῆς ΒΦ 1, τετραπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΝΒ· διπλῆ ἄρα ἐστὶν ἡ ΦΒ τῆς ΒΝ.
And the square on SB is equal to the squares on SN, NB; therefore the squares on BS, SPhi, that is, the square on BPhi, are quadruple of the square on NB; therefore PhiB is double of BN.