§3.prop.26ἐν τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι.
In equal circles equal angles stand on equal circumferences, whether they stand at the centers or at the circumferences.
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.
Let ΑΒΓ, ΔΕΖ be equal circles, and in them let the angles at the centers, ΒΗΓ, ΕΘΖ, and those at the circumferences, ΒΑΓ, ΕΔΖ, be equal; I say that the circumference ΒΚΓ is equal to the circumference ΕΛΖ.
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.
For let ΒΓ, ΕΖ be joined. And since the circles ΑΒΓ, ΔΕΖ are equal, the radii are equal; therefore the two ΒΗ, ΗΓ are equal to the two ΕΘ, ΘΖ; and the angle at Η is equal to the angle at Θ; therefore the base ΒΓ is equal to the base ΕΖ.
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·
And since the angle at Α is equal to the angle at Δ, the segment ΒΑΓ is therefore similar to the segment ΕΔΖ; and they are on equal straight lines.
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.
But similar segments of circles on equal straight lines are equal to one another; therefore the segment ΒΑΓ is equal to ΕΔΖ. But the whole circle ΑΒΓ is also equal to the whole circle ΔΕΖ; therefore the remaining circumference ΒΚΓ is equal to the circumference ΕΛΖ.
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.
Therefore in equal circles equal angles stand on equal circumferences, whether they stand at the centers or at the circumferences; which was to be proved.
§3.prop.27ἐν τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βεβηκυῖαι γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι.
In equal circles the angles standing on equal circumferences are equal to one another, whether they stand at the centers or at the circumferences.
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.
For in equal circles ΑΒΓ, ΔΕΖ on equal circumferences ΒΓ, ΕΖ let the angles ΒΗΓ, ΕΘΖ stand at the centers Η, Θ, and the angles ΒΑΓ, ΕΔΖ at the circumferences; I say that the angle ΒΗΓ is equal to the angle ΕΘΖ, and the angle ΒΑΓ is equal to the angle ΕΔΖ.
εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ τῇ ὑπὸ ΕΘΖ, μία αὐτῶν μείζων ἐστίν.
For if the angle ΒΗΓ is unequal to the angle ΕΘΖ, one of them is greater.
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.
Let the angle ΒΗΓ be greater, and on the straight line ΒΗ and at the point Η on it let the angle ΒΗΚ be constructed equal to the angle ΕΘΖ; but equal angles stand on equal circumferences when they are at the centers; therefore the circumference ΒΚ is equal to the circumference ΕΖ.
ἀλλὰ ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση· καὶ ἡ ΒΚ ἄρα τῇ ΒΓ ἐστιν ἴση ἡ ἐλάττων τῇ μείζονι· ὅπερ ἐστὶν ἀδύνατον.
But ΕΖ is equal to ΒΓ; therefore ΒΚ is also equal to ΒΓ, the less to the greater; which is impossible.
οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ γωνία τῇ ὑπὸ ΕΘΖ· ἴση ἄρα.
Therefore the angle ΒΗΓ is not unequal to the angle ΕΘΖ; therefore it is equal.
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.
And the angle at Α is half of the angle ΒΗΓ, and the angle at Δ is half of the angle ΕΘΖ; therefore the angle at Α is also equal to the angle at Δ. Therefore in equal circles the angles standing on equal circumferences are equal to one another, whether they stand at the centers or at the circumferences; which was to be proved.