In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.

Ἐν τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι. Ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ καὶ ἐν αὐτοῖς ἴσαι γωνίαι ἔστωσαν πρὸς μὲν τοῖς κέντροις αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ: λέγω, ὅτι ἴση ἐστὶν ἡ ΒΚΓ περιφέρεια τῇ ΕΛΖ περιφερείᾳ. Ἐπεζεύχθωσαν γὰρ αἱ ΒΓ, ΕΖ. Καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΑΒΓ, ΔΕΖ κύκλοι, ἴσαι εἰσὶν αἱ ἐκ τῶν κέντρων: δύο δὴ αἱ ΒΗ, ΗΓ δύο ταῖς ΕΘ, ΘΖ ἴσαι: καὶ γωνία ἡ πρὸς τῷ Η γωνίᾳ τῇ πρὸς τῷ Θ ἴση: βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ πρὸς τῷ Α γωνία τῇ πρὸς τῷ Δ, ὅμοιον ἄρα ἐστὶ τὸ ΒΑΓ τμῆμα τῷ ΕΔΖ τμήματι: καί εἰσιν ἐπὶ ἴσων εὐθειῶν [ τῶν ΒΓ, ΕΖ ]: τὰ δὲ ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν: ἴσον ἄρα τὸ ΒΑΓ τμῆμα τῷ ΕΔΖ. ἔστι δὲ καὶ ὅλος ὁ ΑΒΓ κύκλος ὅλῳ τῷ ΔΕΖ κύκλῳ ἴσος: λοιπὴ ἄρα ἡ ΒΚΓ περιφέρεια τῇ ΕΛΖ περιφερείᾳ ἐστὶν ἴση. Ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι: ὅπερ ἔδει δεῖξαι. | In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences. Let ABC, DEF be equal circles, and in them let there be equal angles, namely at the centres the angles BGC, EHF, and at the circumferences the angles BAC, EDF; I say that the circumference BKC is equal to the circumference ELF. For let BC, EF be joined. Now, since the circles ABC, DEF are equal, the radii are equal. Thus the two straight lines BG, GC are equal to the two straight lines EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equal to the base EF. [I. 4] And, since the angle at A is equal to the angle at D, the segment BAC is similar to the segment EDF; [III. Def. 11] and they are upon equal straight lines. But similar segments of circles on equal straight lines are equal to one another; [III. 24] therefore the segment BAC is equal to EDF. But the whole circle ABC is also equal to the whole circle DEF; therefore the circumference BKC which remains is equal to the circumference ELF. |