§3.prop.28ἐν τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι.
In equal circles equal straight lines cut off equal circumferences, the greater equal to the greater and the less to the less.
ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ ἐν τοῖς κύκλοις ἴσαι εὐθεῖαι ἔστωσαν αἱ ΑΒ, ΔΕ τὰς μὲν ΑΓΒ, ΔΖΕ περιφερείας μείζονας ἀφαιροῦσαι τὰς δὲ ΑΗΒ, ΔΘΕ ἐλάττονας· λέγω, ὅτι ἡ μὲν ΑΓΒ μείζων περιφέρεια ἴση ἐστὶ τῇ ΔΖΕ μείζονι περιφερείᾳ, ἡ δὲ ΑΗΒ ἐλάττων περιφέρεια τῇ ΔΘΕ.
εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων τὰ Κ, Λ, καὶ ἐπεζεύχθωσαν αἱ ΑΚ, ΚΒ, ΔΛ, ΛΕ.
καὶ ἐπεὶ ἴσοι κύκλοι εἰσίν, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν κέντρων·
Let ΑΒΓ, ΔΕΖ be equal circles, and in the circles let ΑΒ, ΔΕ be equal straight lines cutting off the greater circumferences ΑΓΒ, ΔΖΕ and the less ΑΗΒ, ΔΘΕ; I say that the greater circumference ΑΓΒ is equal to the greater circumference ΔΖΕ, and the less circumference ΑΗΒ to the less ΔΘΕ.
δύο δὴ αἱ ΑΚ, ΚΒ δυσὶ ταῖς ΔΛ, ΛΕ ἴσαι εἰσίν·
For let the centers of the circles, Κ, Λ, be taken, and let ΑΚ, ΚΒ, ΔΛ, ΛΕ be joined.
καὶ βάσις ἡ ΑΒ βάσει τῇ ΔΕ ἴση· γωνία ἄρα ἡ ὑπὸ ΑΚΒ γωνίᾳ τῇ ὑπὸ ΔΛΕ ἴση ἐστίν.
And since the circles are equal, the radii are also equal; therefore the two ΑΚ, ΚΒ are equal to the two ΔΛ, ΛΕ; and the base ΑΒ is equal to the base ΔΕ; therefore the angle ΑΚΒ is equal to the angle ΔΛΕ.
αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ὅταν πρὸς τοῖς κέντροις ὦσιν· ἴση ἄρα ἡ ΑΗΒ περιφέρεια τῇ ΔΘΕ. ἐστὶ δὲ καὶ ὅλος ὁ ΑΒΓ κύκλος ὅλῳ τῷ ΔΕΖ κύκλῳ ἴσος·
But equal angles stand on equal circumferences, when they are at the centers; therefore the circumference ΑΗΒ is equal to the circumference ΔΘΕ.
καὶ λοιπὴ ἄρα ἡ ΑΓΒ περιφέρεια λοιπῇ τῇ ΔΖΕ περιφερείᾳ ἴση ἐστίν.
And the whole circle ΑΒΓ is also equal to the whole circle ΔΕΖ; therefore the remaining circumference ΑΓΒ is equal to the remaining circumference ΔΖΕ.
ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι· ὅπερ ἔδει δεῖξαι.
Therefore in equal circles equal straight lines cut off equal circumferences, the greater to the greater and the less to the less; which was to be proved.
§3.prop.29ἐν τοῖς ἴσοις κύκλοις τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν.
In equal circles equal straight lines subtend equal circumferences.
ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ ἐν αὐτοῖς ἴσαι περιφέρειαι ἀπειλήφθωσαν αἱ ΒΗΓ, ΕΘΖ, καὶ ἐπεζεύχθωσαν αἱ ΒΓ, ΕΖ εὐθεῖαι· λέγω, ὅτι ἴση ἐστὶν ἡ ΒΓ τῇ ΕΖ.
εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων, καὶ ἔστω τὰ Κ, Λ, καὶ ἐπεζεύχθωσαν αἱ ΒΚ, ΚΓ, ΕΛ, ΛΖ.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΗΓ περιφέρεια τῇ ΕΘΖ περιφερείᾳ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΚΓ τῇ ὑπὸ ΕΛΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΑΒΓ, ΔΕΖ κύκλοι, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν κέντρων·
Let ΑΒΓ, ΔΕΖ be equal circles, and in them let equal circumferences ΒΗΓ, ΕΘΖ be cut off, and let the straight lines ΒΓ, ΕΖ be joined; I say that ΒΓ is equal to ΕΖ.
δύο δὴ αἱ ΒΚ, ΚΓ δυσὶ ταῖς ΕΛ, ΛΖ ἴσαι εἰσίν·
For let the centers of the circles be taken, and let them be Κ, Λ, and let ΒΚ, ΚΓ, ΕΛ, ΛΖ be joined.
καὶ γωνίας ἴσας περιέχουσιν·
And since the circumference ΒΗΓ is equal to the circumference ΕΘΖ, the angle ΒΚΓ is also equal to the angle ΕΛΖ.
βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν.
And since the circles ΑΒΓ, ΔΕΖ are equal, the radii are also equal; therefore the two ΒΚ, ΚΓ are equal to the two ΕΛ, ΛΖ; and they contain equal angles; therefore the base ΒΓ is equal to the base ΕΖ.
ἐν ἄρα τοῖς ἴσοις κύκλοις τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν· ὅπερ ἔδει δεῖξαι.
Therefore in equal circles equal straight lines subtend equal circumferences; which was to be proved.
§3.prop.30τὴν δοθεῖσαν περιφέρειαν δίχα τεμεῖν.
To bisect a given circumference.
ἔστω ἡ δοθεῖσα περιφέρεια ἡ ΑΔΒ· δεῖ δὴ τὴν ΑΔΒ περιφέρειαν δίχα τεμεῖν.
Let ΑΔΒ be the given circumference; it is required to bisect the circumference ΑΔΒ.
ἐπεζεύχθω ἡ ΑΒ, καὶ τετμήσθω δίχα κατὰ τὸ Γ, καὶ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ ΓΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΒ.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, κοινὴ δὲ ἡ ΓΔ, δύο δὴ αἱ ΑΓ, ΓΔ δυσὶ ταῖς ΒΓ, ΓΔ ἴσαι εἰσίν·
Let ΑΒ be joined, and let it be bisected at Γ, and from the point Γ let ΓΔ be drawn at right angles to the straight line ΑΒ, and let ΑΔ, ΔΒ be joined.
καὶ γωνία ἡ ὑπὸ ΑΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση· ὀρθὴ γὰρ ἑκατέρα· βάσις ἄρα ἡ ΑΔ βάσει τῇ ΔΒ ἴση ἐστίν.
And since ΑΓ is equal to ΓΒ, and ΓΔ is common, the two ΑΓ, ΓΔ are equal to the two ΒΓ, ΓΔ; and the angle ΑΓΔ is equal to the angle ΒΓΔ, for each is a right angle; therefore the base ΑΔ is equal to the base ΔΒ.
αἱ δὲ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι· καί ἐστιν ἑκατέρα τῶν ΑΔ, ΔΒ περιφερειῶν ἐλάττων ἡμικυκλίου· ἴση ἄρα ἡ ΑΔ περιφέρεια τῇ ΔΒ περιφερείᾳ.
But equal straight lines cut off equal circumferences, the greater to the greater and the less to the less; and each of the circumferences ΑΔ, ΔΒ is less than a semicircle; therefore the circumference ΑΔ is equal to the circumference ΔΒ.
ἡ ἄρα δοθεῖσα περιφέρεια δίχα τέτμηται κατὰ τὸ Δ σημεῖον· ὅπερ ἔδει ποιῆσαι.
Therefore the given circumference has been bisected at the point Δ; which was to be done.