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In equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences.
| Ἐν τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βεβηκυῖαι γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι. Ἐν γὰρ ἴσοις κύκλοις τοῖς ΑΒΓ, ΔΕΖ ἐπὶ ἴσων περιφερειῶν τῶν ΒΓ, ΕΖ πρὸς μὲν τοῖς Η, Θ κέντροις γωνίαι βεβηκέτωσαν αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ: λέγω, ὅτι ἡ μὲν ὑπὸ ΒΗΓ γωνία τῇ ὑπὸ ΕΘΖ ἐστιν ἴση, ἡ δὲ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. Εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ τῇ ὑπὸ ΕΘΖ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΒΗΓ, καὶ συνεστάτω πρὸς τῇ ΒΗ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Η τῇ ὑπὸ ΕΘΖ γωνίᾳ ἴση ἡ ὑπὸ ΒΗΚ: αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ὅταν πρὸς τοῖς κέντροις ὦσιν: ἴση ἄρα ἡ ΒΚ περιφέρεια τῇ ΕΖ περιφερείᾳ. ἀλλὰ ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση: καὶ ἡ ΒΚ ἄρα τῇ ΒΓ ἐστιν ἴση ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ γωνία τῇ ὑπὸ ΕΘΖ: ἴση ἄρα. καί ἐστι τῆς μὲν ὑπὸ ΒΗΓ ἡμίσεια ἡ πρὸς τῷ Α, τῆς δὲ ὑπὸ ΕΘΖ ἡμίσεια ἡ πρὸς τῷ Δ: ἴση ἄρα καὶ ἡ πρὸς τῷ Α γωνία τῇ πρὸς τῷ Δ. Ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βεβηκυῖαι γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι: ὅπερ ἔδει δεῖξαι. | In equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences. For in equal circles ABC, DEF, on equal circumferences BC, EF, let the angles BGC, EHF stand at the centres G, H, and the angles BAC, EDF at the circumferences; I say that the angle BGC is equal to the angle EHF, and the angle BAC is equal to the angle EDF. For, if the angle BGC is unequal to the angle EHF, one of them is greater. Let the angle BGC be greater : and on the straight line BG, and at the point G on it, let the angle BGK be constructed equal to the angle EHF. [I. 23] Now equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference BK is equal to the circumference EF. But EF is equal to BC; therefore BK is also equal to BC, the less to the greater : which is impossible. Therefore the angle BGC is not unequal to the angle EHF; therefore it is equal to it. And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF; [III. 20] therefore the angle at A is also equal to the angle at D. |