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The opposite angles of quadrilaterals in circles are equal to two right angles.
| Τῶν ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. Ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ τετράπλευρον ἔστω τὸ ΑΒΓΔ: λέγω, ὅτι αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. Ἐπεζεύχθωσαν αἱ ΑΓ, ΒΔ. Ἐπεὶ οὖν παντὸς τριγώνου αἱ τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν, τοῦ ΑΒΓ ἄρα τριγώνου αἱ τρεῖς γωνίαι αἱ ὑπὸ ΓΑΒ, ΑΒΓ, ΒΓΑ δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἴση δὲ ἡ μὲν ὑπὸ ΓΑΒ τῇ ὑπὸ ΒΔΓ: ἐν γὰρ τῷ αὐτῷ τμήματί εἰσι τῷ ΒΑΔΓ: ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΑΔΒ: ἐν γὰρ τῷ αὐτῷ τμήματί εἰσι τῷ ΑΔΓΒ: ὅλη ἄρα ἡ ὑπὸ ΑΔΓ ταῖς ὑπὸ ΒΑΓ, ΑΓΒ ἴση ἐστίν. κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ: αἱ ἄρα ὑπὸ ΑΒΓ, ΒΑΓ, ΑΓΒ ταῖς ὑπὸ ΑΒΓ, ΑΔΓ ἴσαι εἰσίν. ἀλλ' αἱ ὑπὸ ΑΒΓ, ΒΑΓ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. καὶ αἱ ὑπὸ ΑΒΓ, ΑΔΓ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ ΒΑΔ, ΔΓΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. Τῶν ἄρα ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν: ὅπερ ἔδει δεῖξαι. | The opposite angles of quadrilaterals in circles are equal to two right angles. Let ABCD be a circle, and let ABCD be a quadrilateral in it; I say that the opposite angles are equal to two right angles. Let AC, BD be joined. Then, since in any triangle the three angles are equal to two right angles, [I. 32] the three angles CAB, ABC, BCA of the triangle ABC are equal to two right angles. But the angle CAB is equal to the angle BDC, for they are in the same segment BADC; [III. 21] and the angle ACB is equal to the angle ADB, for they are in the same segment ADCB; therefore the whole angle ADC is equal to the angles BAC, ACB. Let the angle ABC be added to each; therefore the angles ABC, BAC, ACB are equal to the angles ABC, ADC. But the angles ABC, BAC, ACB are equal to two right angles; therefore the angles ABC, ADC are also equal to two right angles. Similarly we can prove that the angles BAD, DCB are also equal to two right angles. |