§13.prop.13#2Ἐκβεβλήσθω γὰρ ἐπʼ εὐθείας τῇ ΚΘ εὐθεῖα ἡ ΘΛ, καὶ κείσθω τῇ ΓΒ ἴση ἡ ΘΛ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΓ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΔ πρὸς τὴν ΓΒ, ἴση δὲ ἡ μὲν ΑΓ τῇ ΚΘ, ἡ δὲ ΓΔ τῇ ΘΕ, ἡ δὲ ΓΒ τῇ ΘΛ, ἔστιν ἄρα ὡς ἡ ΚΘ πρὸς τὴν ΘΕ, οὕτως ἡ ΕΘ πρὸς τὴν ΘΛ·
let the straight line ThL be produced in a straight line with KTh, and let ThL be made equal to GB. And since, as AC is to GD, so is GD to GB, and AC is equal to KTh, and GD to ThE, and GB to ThL, therefore as KTh is to ThE, so is ETh to ThL.
τὸ ἄρα ὑπὸ τῶν ΚΘ, ΘΛ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ. καί ἐστιν ὀρθὴ ἑκατέρα τῶν ὑπὸ ΚΘΕ, ΕΘΛ γωνιῶν·
Therefore the rectangle contained by KTh, ThL is equal to the square on ETh. And each of the angles KThE, EThL is a right angle.
τὸ ἄρα ἐπὶ τῆς ΚΛ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Ε.
Therefore the semicircle described on KL will also pass through E.
ἐὰν δὴ μενούσης τῆς ΚΛ περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, ἥξει καὶ διὰ τῶν Ζ, Η σημείων ἐπιζευγνυμένων τῶν ΖΛ, ΛΗ καὶ ὀρθῶν ὁμοίως γινομένων τῶν πρὸς τοῖς Ζ, Η γωνιῶν·
If then, while KL remains fixed, the semicircle be carried round and restored again to the same position from which it began to be moved, it will also pass through the points Z, H, if ZL, LH are joined and the angles at Z, H similarly become right angles.
καὶ ἔσται ἡ πυραμὶς σφαίρᾳ περιειλημμένη τῇ δοθείσῃ.
And the pyramid will have been comprehended in the given sphere.
ἡ γὰρ ΚΛ τῆς σφαίρας διάμετρος ἴση ἐστὶ τῇ τῆς δοθείσης σφαίρας διαμέτρῳ τῇ ΑΒ, ἐπειδήπερ τῇ μὲν ΑΓ ἴση κεῖται ἡ ΚΘ, τῇ δὲ ΓΒ ἡ ΘΛ.
λέγω δή, ὅτι ἡ τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ δυνάμει τῆς πλευρᾶς τῆς πυραμίδος.
For the diameter KL of the sphere is equal to the diameter AB of the given sphere, since indeed KTh is laid down equal to AG, and ThL to GB. I say then, that the diameter of the sphere is one and a half times in square of the side of the pyramid.
ἐπεὶ γὰρ διπλῆ ἐστιν ἡ ΑΓ τῆς ΓΒ, τριπλῆ ἄρα ἐστὶν ἡ ΑΒ τῆς ΒΓ· ἀναστρέψαντι ἡμιολία ἄρα ἐστὶν ἡ ΒΑ τῆς ΑΓ. ὡς δὲ ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΔ.
For since AG is double of GB, therefore AB is triple of BG; therefore, by conversion, BA is one and a half times of AG. And as BA is to AG, so is the square on BA to the square on AD.
ἡμιόλιον ἄρα καὶ τὸ ἀπὸ τῆς ΒΑ τοῦ ἀπὸ τῆς ΑΔ. καί ἐστιν ἡ μὲν ΒΑ ἡ τῆς δοθείσης σφαίρας διάμετρος, ἡ δὲ ΑΔ ἴση τῇ πλευρᾷ τῆς πυραμίδος.
Therefore the square on BA is also one and a half times of the square on AD. And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid.
ἡ ἄρα τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ τῆς πλευρᾶς τῆς πυραμίδος· ὅπερ ἔδει δεῖξαι.
Therefore the diameter of the sphere is one and a half times in square of the side of the pyramid; which it was required to prove.
λῆμμα
δεικτέον, ὅτι ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ.
Ἐκκείσθω γὰρ ἡ τοῦ ἡμικυκλίου καταγραφή, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΑΓ τετράγωνον τὸ ΕΓ, καὶ συμπεπληρώσθω τὸ ΖΒ παραλληλόγραμμον.
Lemma It is to be proved that, as AB is to BC, so is the square on AD to the square on DG. For let the figure of the semicircle be set out, and let DB be joined, and let the square EG be described on AC, and let the parallelogram ZB be completed.
ἐπεὶ οὖν διὰ τὸ ἰσογώνιον εἶναι τὸ ΔΑΒ τρίγωνον τῷ ΔΑΓ τριγώνῳ ἐστὶν ὡς ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΔΑ πρὸς τὴν ΑΓ, τὸ ἄρα ὑπὸ τῶν ΒΑ, ΑΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΔ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ΕΒ πρὸς τὸ ΒΖ, καί ἐστι τὸ μὲν ΕΒ τὸ ὑπὸ τῶν ΒΑ, ΑΓ·
Since then, because the triangle DAB is equiangular with the triangle DAG, as BA is to AD, so is DA to AC, therefore the rectangle contained by BA, AC is equal to the square on AD.
ἴση γὰρ ἡ ΕΑ τῇ ΑΓ· τὸ δὲ ΒΖ τὸ ὑπὸ τῶν ΑΓ, ΓΒ, ὡς ἄρα ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ὑπὸ τῶν ΒΑ, ΑΓ πρὸς τὸ ὑπὸ τῶν ΑΓ, ΓΒ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΒΑ, ΑΓ ἴσον τῷ ἀπὸ τῆς ΑΔ, τὸ δὲ ὑπὸ τῶν ΑΓΒ ἴσον τῷ ἀπὸ τῆς ΔΓ·
And since, as AB is to BC, so is EB to BZ, and EB is the rectangle contained by BA, AC (for EA is equal to AC), and BZ is the rectangle contained by AC, GB, therefore as AB is to BC, so is the rectangle contained by BA, AC to the rectangle contained by AC, GB.
ἡ γὰρ ΔΓ κάθετος τῶν τῆς βάσεως τμημάτων τῶν ΑΓ, ΓΒ μέση ἀνάλογόν ἐστι διὰ τὸ ὀρθὴν εἶναι τὴν ὑπὸ ΑΔΒ. ὡς ἄρα ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΓ· ὅπερ ἔδει δεῖξαι.
And the rectangle contained by BA, AC is equal to the square on AD, and the rectangle contained by AC, GB is equal to the square on DG (for the perpendicular DG is a mean proportional between the segments AC, GB of the base, because the angle ADB is a right angle). Therefore, as AB is to BC, so is the square on AD to the square on DG; which it was required to prove.