Page 444 - Euclid's Elements of Geometry

ST EW iaþ.

ELEMENTS BOOK 11

λοιπὴ ἄρα ἡ ΔΓ λοιπῆς τῆς ΕΓ μείζων ἐστίν. καὶ ἐπεὶ ἴση the two (straight-lines AD and AB are) equal to the
ἐστὶν ἡ ΔΑ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΓ, καὶ βάσις ἡ ΔΓ βάσεως two (straight-lines EA and AB, respectively). And an-
τῆς ΕΓ μείζων ἐστίν, γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνάις τῆς ὑπὸ gle DAB (is) equal to angle BAE. Thus, the base DB
ΕΑΓ μείζων ἐστίν. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΔΑΒ τῇ ὑπὸ ΒΑΕ is equal to the base BE [Prop. 1.4]. And since the (sum
ἴση· αἱ ἄρα ὑπὸ ΔΑΒ, ΔΑΓ τῆς ὑπὸ ΒΑΓ μείζονές εἰσιν. of the) two (straight-lines) BD and DC is greater than
ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ λοιπαὶ σύνδυο λαμβανόμεναι BC [Prop. 1.20], of which DB was shown (to be) equal
τῆς λοιπῆς μείζονές εἰσιν. to BE, the remainder DC is thus greater than the re-
᾿Εὰν ἄρα στερεὰ γωνία ὑπὸ τριῶν γωνιῶν ἐπιπέδων mainder EC. And since DA is equal to AE, but AC
περιέχηται, δύο ὁποιαιοῦν τῆς λοιπῆς μείζονές εἰσι πάντῃ (is) common, and the base DC is greater than the base
μεταλαμβανόμεναι· ὅπερ ἔδει δεῖξαι. EC, the angle DAC is thus greater than the angle EAC
[Prop. 1.25]. And DAB was also shown (to be) equal to
BAE. Thus, (the sum of) DAB and DAC is greater than
kaþ remaining (one).
BAC. So, similarly, we can also show that the remain-
ing (angles), being taken in pairs, are greater than the

Thus, if a solid angle is contained by three plane an-
gles then (the sum of) any two (angles) is greater than
the remaining (one), (the angles) being taken up in any
(possible way). (Which is) the very thing it was required
to show.
G . Proposition 21

῞Απασα στερεὰ γωνία ὑπὸ ἐλασσόνων [ἢ] τεσσάρων Any solid angle is contained by plane angles (whose
ὀρθῶν γωνιῶν ἐπιπέδων περιέχεται. sum is) less [than] four right-angles. †
C

B A D A D

B
῎Εστω στερεὰ γωνία ἡ πρὸς τῷ Α περιεχομένη ὑπὸ Let the solid angle A be contained by the plane angles
ἐπιπέδων γωνιῶν τῶν ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ· λέγω, ὅτι αἱ BAC, CAD, and DAB. I say that (the sum of) BAC,
ὑπὸ ΒΑΓ, ΓΑΔ, ΔΑΒ τεσσάρων ὀρθῶν ἐλάσσονές εἰσιν. CAD, and DAB is less than four right-angles.
Εἰλήφθω γὰρ ἐφ᾿ ἑκάστης τῶν ΑΒ, ΑΓ, ΑΔ τυχόντα For let the random points B, C, and D have been
σημεῖα τὰ Β, Γ, Δ, καὶ ἐπεζεύχθωσαν αἱ ΒΓ, ΓΔ, ΔΒ. καὶ taken on each of (the straight-lines) AB, AC, and AD
ἐπεὶ στερεὰ γωνία ἡ πρὸς τῷ Β ὑπὸ τριῶν γωνιῶν ἐπιπέδων (respectively). And let BC, CD, and DB have been
περιέχεται τῶν ὑπὸ ΓΒΑ, ΑΒΔ, ΓΒΔ, δύο ὁποιαιοῦν τῆς joined. And since the solid angle at B is contained
λοιπῆς μείζονές εἰσιν· αἱ ἄρα ὑπὸ ΓΒΑ, ΑΒΔ τῆς ὑπὸ ΓΒΔ by the three plane angles CBA, ABD, and CBD, (the
μείζονές εἰσιν. διὰ τὰ αὐτὰ δὴ καὶ αἱ μὲν ὑπὸ ΒΓΑ, ΑΓΔ sum of) any two is greater than the remaining (one)
τῆς ὑπὸ ΒΓΔ μείζονές εἰσιν, αἱ δὲ ὑπὸ ΓΔΑ, ΑΔΒ τῆς ὑπὸ [Prop. 11.20]. Thus, (the sum of) CBA and ABD is
ΓΔΒ μείζονές εἰσιν· αἱ ἓξ ἄρα γωνίαι αἱ ὑπὸ ΓΒΑ, ΑΒΔ, greater than CBD. So, for the same (reasons), (the sum
ΒΓΑ, ΑΓΔ, ΓΔΑ, ΑΔΒ τριῶν τῶν ὑπὸ ΓΒΔ, ΒΓΑ, ΓΔΒ of) BCA and ACD is also greater than BCD, and (the
μείζονές εἰσιν. ἀλλὰ αἱ τρεῖς αἱ ὑπὸ ΓΒΔ, ΒΔΓ, ΒΓΔ δυσὶν sum of) CDA and ADB is greater than CDB. Thus,
ὀρθαῖς ἴσαι εἰσίν· αἱ ἓξ ἄρα αἱ ὑπὸ ΓΒΑ, ΑΒΔ, ΒΓΑ, ΑΓΔ, the (sum of the) six angles CBA, ABD, BCA, ACD,
ΓΔΑ, ΑΔΒ δύο ὀρθῶν μείζονές εἰσιν. καὶ ἐπεὶ ἑκάστου τῶν CDA, and ADB is greater than the (sum of the) three
ΑΒΓ, ΑΓΔ, ΑΔΒ τριγώνων αἱ τρεῖς γωνίαι δυσὶν ὀρθαῖς (angles) CBD, BCD, and CDB. But, the (sum of the)
ἴσαι εἰσίν, αἱ ἄρα τῶν τριῶν τριγώνων ἐννέα γωνίαι αἱ ὑπὸ three (angles) CBD, BDC, and BCD is equal to two

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