§1#3μὲν γᾶς τὸ Θ, τοῦ δὲ ἁλίου τὸ Κ, ὄψις δὲ ἔστω τὸ △, καὶ ἄχθωσαν εὐθεῖαι ἐπιψαύουσαι τοῦ ΣΗ κύκλου ἀπὸ μὲν τοῦ △ αἱ △Λ, △Ξ, ἐπιψαυόντων δὲ κατὰ τὸ Ν καὶ τὸ Τ, ἀπὸ δὲ τοῦ Θ αἱ ΘΜ, ΘΟ, ἐπιψαυόντων δὲ κατὰ τὸ Χ καὶ τὸ Ρ, τὸν δὲ ΑΒΓ κύκλον τεμνόντων αἱ ΘΜ, ΘΟ κατὰ τὸ Α καὶ τὸ Β ἔστι δὴ μείζων ἁ ΘΚ τᾶς △Κ, ἐπεὶ ὑπόκειται ὁ ἅλιος ὑπὲρ τὸν ὁρίζοντα εἶμεν· ὥστε ἁ γωνία ἁ περιεχομένα ὑπὸ τᾶν △Λ, △Ξ μείζων ἐστὶ τᾶς γωνίας τᾶς περιεχομένας ὑπὸ τᾶν ΘΜ, ΘΟ, Ἁ δὲ περιεχομένα γωνία ὑπὸ τᾶν △Λ, △Ξ μείζων μὲν ἐστιν ἢ διακοσιοστὸν μέρος ὀρθᾶς, ἐλάττων δὲ ἢ τᾶς ὀρθᾶς διαιρεθείσας εἰς ρξδ τούτων ἓν μέρος· ἴσα γάρ ἐστιν τᾷ γωνίᾳ, εἰς ἃν ὁ ἅλιος ἐναρμόζει τὰν κορυφὰν ἔχουσαν ποτὶ τᾷ ὄψει ὥστε ἁ γωνία ἁ περιεχομένα ὑπὸ τᾶν ΘΜ, ΘΟ ἐλάττων ἐστὶν ἢ τᾶς ὀρθᾶς διαιρεθείσας εἰς ρξδ τούτων ἓν μέρος, ἁ δὲ ΑΒ εὐθεῖα ἐλάττων ἐστὶ τᾶς ὑποτεινούσας ἓν τμᾶμα διαιρεθείσας τᾶς τοῦ ΑΒΓ κύκλου περιφερείας ἐς χνς΄.
Let Θ be the center of the earth, Κ the center of the sun, △ the eye, and let there be drawn straight lines touching the circle ΣΗ, namely △Λ, △Ξ from △, touching at Ν and Τ, and ΘΜ, ΘΟ from Θ, touching at Χ and Ρ, and let ΘΜ, ΘΟ cut the circle ΑΒΓ at Α and Β. Now, ΘΚ is greater than △Κ, since it is assumed that the sun is above the horizon; so that the angle contained by △Λ, △Ξ is greater than the angle contained by ΘΜ, ΘΟ. And the angle contained by △Λ, △Ξ is greater than one two-hundredth part of a right angle, but less than one part of a right angle divided into 164; for it is equal to the angle which the sun fits, having its vertex at the eye; so that the angle contained by ΘΜ, ΘΟ is less than one part of a right angle divided into 164, and the straight line ΑΒ is less than the chord subtending one arc of the circumference of the circle ΑΒΓ divided into 656.
Α δὲ τοῦ εἰρημένου πολυγωνίου περίμετρος ποτὶ τὰν ἐκ τοῦ κέντρου τοῦ ΑΒΓ κύκλου ἐλάττονα λόγον ἔχει ἢ τὰ μδ ποτὶ τὰ ζ διὰ τὸ παντὸς πολυγωνίου ἐγγεγραμμένου ἐν κύκλῳ τὰν περίμετρον ποτὶ τὰν ἐκ τοῦ κέντρου ἐλάττονα λόγον ἔχειν ἢ τὰ μδ ποτὶ τὰ ζ· ἐπίστασαι γὰρ δεδειγμένον ὑφʼ ἁμῶν ὅτι παντὸς κύκλου ἁ περιφέρεια μείζων ἐστὶν ἢ τριπλασίων τᾶς διαμέτρου ἐλάσσονι ἢ ἑβδόμῳ μέρει, ταύτας δὲ ἐλάττων ἐστὶν ἁ περίμετρος τοῦ ἐγγραφέντος πολυγωνίου· ἐλάττω οὖν λόγον ἔχει ἁ ΒΑ ποτὶ τὰν ΘΚ ἢ τὰ ια ποτὶ τὰ αρμη·
And the perimeter of the said polygon has to the radius of the circle ΑΒΓ a less ratio than 44 has to 7, because the perimeter of any polygon inscribed in a circle has to the radius a less ratio than 44 has to 7; for you know that it has been proved by us that the circumference of any circle is greater than three times the diameter by an amount less than one-seventh part, and the perimeter of the inscribed polygon is less than this circumference; therefore, ΒΑ has to ΘΚ a less ratio than 11 has to 1148; so that ΒΑ is less than one-hundredth part of ΘΚ.
ὥστε ἐλάττων ἐστὶν ἁ ΒΑ τᾶς ΘΚ ἢ ἑκατοστὸν μέρος. Τᾷ δὲ ΒΑ ἴσα ἐστὶν ἁ διάμετρος τοῦ ΣΗ κύκλου, διότι καὶ ἁ ἡμίσεια αὐτᾶς ἁ ΦΑ ἴσα ἐστὶ τᾷ ΚΡ ἰσᾶν γὰρ ἐουσᾶν τᾶν ΘΚ, ΘΑ ἀπὸ τῶν περάτων κάθετοι ἐπιζεύγνυνται ὑπὸ τὰν αὐτὰν γωνίαν· δῆλον οὖν ὅτι ἁ διάμετρος τοῦ ΣΗ κύκλου ἐλάττων ἐστὶν ἢ ἑκατοστὸν μέρος τᾶς ΘΚ. Καὶ ἁ ΕΘΥ διάμετρος ἐλάττων ἐστὶ τᾶς διαμέτρου τοῦ ΣΗ κύκλου, ἐπεὶ ἐλάττων ἐστὶν ὁ △ΕΖ κύκλος τοῦ ΣΗ κύκλου· ἐλάττονες ἄρα ἐντὶ ἀμφότεραι αἱ ΘΥ, ΚΣ ἢ ἑκατοστὸν μέρος τᾶς ΘΚ· ὥστε ἁ ΘΚ ποτὶ τὰν ΥΣ ἐλάττονά λόγον ἔχει ἢ τὰ ρ ποτὶ τὰ (??)θ.
And the diameter of the circle ΣΗ is equal to ΒΑ, because its half ΦΑ is equal to ΚΡ; for, since ΘΚ, ΘΑ are equal, perpendiculars are drawn from their extremities under the same angle; it is therefore clear that the diameter of the circle ΣΗ is less than one-hundredth part of ΘΚ. And the diameter ΕΘΥ is less than the diameter of the circle ΣΗ, since the circle △ΕΖ is less than the circle ΣΗ; therefore, both ΘΥ, ΚΣ together are less than one-hundredth part of ΘΚ; so that ΘΚ has to ΥΣ a less ratio than 100 has to 99.
Καὶ ἐπεὶ ἁ μὲν ΘΚ οὐκ ἐλάττων ἐστὶ τᾶς ΘΡ, ἁ δὲ ΣΥ ἐλάττων τᾶς △Τ, ἐλάττω ἄρα κα λόγον ἔχοι ἁ ΘΡ ποτὶ τὰν △Τ ἢ τὰ ρ ποτὶ τὸ (??)θ.
And since ΘΚ is not less than ΘΡ, and ΣΥ is less than △Τ, ΘΡ would therefore have a less ratio to △Τ than 100 has to 99.
Ἐπεὶ δὲ τῶν ΘΚΡ, △ΚΤ ὀρθογωνίων ἐόντων αἱ μὲν ΚΡ, ΚΤ πλευραὶ ἴσαι ἐντὶ, αἱ δὲ ΘΡ, △Τ ἀνίσοι καὶ μείζων ἁ ΘΡ, ἁ γωνία ἁ περιεχομένα ὑπὸ τᾶν △Τ, △Κ ποτὶ τὰν γωνίαν τὰν περιεχομέναν ὑπὸ τᾶν ΘΡ, ΘΚ μείζονα μὲν ἔχει λόγον ἢ ἁ ΘΚ ποτὶ τὰν △Κ, ἐλάττω δὲ ἢ ἁ ΘΡ ποτὶ τὰν △Τ εἰ γάρ κα δυῶν τριγώνων ὀρθογωνίων αἱ μὲν ἅτεραι πλευραὶ αἱ περὶ τὰν ὀρθὰν γωνίαν ἴσαι ἔωντι, αἱ δὲ ἅτεραι ἄνισοι, ἁ μείζων γωνία τᾶν ποτὶ ταῖς ἀνίσοις πλευραῖς ποτὶ τὰν ἐλάττονα μείζονα μὲν ἔχει λόγον ἢ ἁ μείζων γραμμὰ τᾶν ὑπὸ τὰν ὀρθὰν γωνίαν ὑποτεινουσᾶν ποτὶ τὰν ἐλάττονα, ἐλάττονα δὲ ἢ ἁ μείζων γραμμὰ τᾶν περὶ τὰν ὀρθὰν γωνίαν ποτὶ τὰν ἐλάττονα.
And since, in the right-angled triangles ΘΚΡ, △ΚΤ, the sides ΚΡ, ΚΤ are equal, and the sides ΘΡ, △Τ are unequal and ΘΡ is greater, the angle contained by △Τ, △Κ has to the angle contained by ΘΡ, ΘΚ a greater ratio than ΘΚ has to △Κ, but a less ratio than ΘΡ has to △Τ; for if in two right-angled triangles one of the sides about the right angle are equal to one another, and the other sides are unequal, the greater angle of those adjacent to the unequal sides has to the less angle a greater ratio than the greater of the lines subtending the right angle has to the less, but a less ratio than the greater of the sides about the right angle has to the less.
Ὥστε ἁ γωνία ἁ περιεχομένα ὑπὸ τᾶν △Λ, △Ξ ποτὶ τὰν γωνίαν τὰν περιεχομέναν ὑπὸ τᾶν ΘΟ, ΘΜ ἐλάττω λόγον ἔχει ἢ ἁ ΘΡ ποτὶ τὰν △Τ, ἅτις ἐλάττω λόγον ἔχει ἢ τὰ ρ ποτὶ τὰ (??)θ· ὥστε καὶ ἁ γωνία ἁ περιεχομένα ὑπὸ τᾶν △Λ, △Ξ ποτὶ τὰν γωνίαν τὰν περιεχομέναν ὑπὸ τᾶν ΘΜ, ΘΟ ἐλάττω λόγον ἔχει ἢ τὰ ρ ποτὶ τὰ (??)θ.
Therefore, the angle contained by △Λ, △Ξ has to the angle contained by ΘΟ, ΘΜ a less ratio than ΘΡ has to △Τ, which has a less ratio than 100 has to 99; so that also the angle contained by △Λ, △Ξ has to the angle contained by ΘΜ, ΘΟ a less ratio than 100 has to 99.
Καὶ ἐπεί ἐστιν ἁ γωνία ἁ περιεχομένα ὑπὸ τᾶν △Λ, △Ξ μείζων ἢ διακοσιοστὸν μέρος ὀρθᾶς, εἴη κα ἁ γωνία ἁ περιεχομένα ὑπὸ τᾶν ΘΜ, ΘΟ μείζων ἢ τᾶς ὀρθᾶς διαιρεθείσας ἐς δισμύρια τούτων (??)θ μέρεα ὥστε μείζων ἐστὶν ἢ διαιρεθείσας τᾶς ὀρθᾶς εἰς σ καὶ γ τούτων ἓν μέρος.
And since the angle contained by △Λ, △Ξ is greater than one two-hundredth part of a right angle, the angle contained by ΘΜ, ΘΟ would be greater than 99 parts of a right angle divided into 20,000; so that it is greater than one part of a right angle divided into 203.
Ἁ ἄρα ΒΑ μείζων ἐστὶ τᾶς ὑποτεινούσας ἓν τμᾶμα διῃρημένας τᾶς τοῦ ΑΒΓ κύκλου περιφερείας εἰς ωιβ.
Therefore, ΒΑ is greater than the chord subtending one arc of the circumference of the circle ΑΒΓ divided into 812.
Τᾷ δὲ ΑΒ ἴσα ἐντὶ ἁ τοῦ ἁλίου διάμετρος δῆλον οὖν ὅτι μείζων ἐστὶν ἁ τοῦ ἁλίου διάμετρος τᾶς τοῦ χιλιαγώνου πλευρᾶς.
And the diameter of the sun is equal to ΑΒ; it is therefore clear that the diameter of the sun is greater than the side of the chiliagon.