§3.prop.36ἐὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπʼ αὐτοῦ πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφαπτομένης τετραγώνῳ.
If any point is taken outside a circle, and from it two straight lines fall on the circle, and one of them cuts the circle, and the other touches it, the rectangle contained by the whole of the cutting line and the part of it cut off outside between the point and the convex circumference will be equal to the square on the tangent.
κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο εὐθεῖαι αἱ ΔΓ, ΔΒ· καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν ΑΒΓ κύκλον, ἡ δὲ ΒΔ ἐφαπτέσθω· λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ τετραγώνῳ.
For let any point Δ be taken outside the circle ΑΒΓ, and from Δ let two straight lines ΔΓ, ΔΒ fall on the circle ΑΒΓ; and let ΔΓΑ cut the circle ΑΒΓ, and let ΒΔ touch it; I say that the rectangle contained by ΑΔ, ΔΓ is equal to the square on ΔΒ.
ἡ ἄρα ΓΑ ἤτοι διὰ τοῦ κέντρου ἐστὶν ἢ οὔ.
Therefore ΓΑ is either through the center or not.
ἔστω πρότερον διὰ τοῦ κέντρου, καὶ ἔστω τὸ Ζ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἐπεζεύχθω ἡ ΖΒ·
First, let it be through the center, and let Ζ be the center of the circle ΑΒΓ, and let ΖΒ be joined; therefore the angle ΖΒΔ is right.
ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΒΔ. καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ δίχα τέτμηται κατὰ τὸ Ζ, πρόσκειται δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ·
And since the straight line ΑΓ has been bisected at Ζ, and ΓΔ is added to it, therefore the rectangle contained by ΑΔ, ΔΓ together with the square on ΖΓ is equal to the square on ΖΔ.
τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. τῷ δὲ ἀπὸ τῆς ΖΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΖΒ, ΒΔ·
And ΖΓ is equal to ΖΒ; therefore the rectangle contained by ΑΔ, ΔΓ together with the square on ΖΒ is equal to the square on ΖΔ.
τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΒ, ΒΔ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΖΒ·
But the squares on ΖΒ, ΒΔ are equal to the square on ΖΔ; therefore the rectangle contained by ΑΔ, ΔΓ together with the square on ΖΒ is equal to the squares on ΖΒ, ΒΔ.
λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ ἐφαπτομένης.
Let the square on ΖΒ be subtracted from each; therefore the remaining rectangle contained by ΑΔ, ΔΓ is equal to the square on the tangent ΔΒ.
ἀλλὰ δὴ ἡ ΔΓΑ μὴ ἔστω διὰ τοῦ κέντρου τοῦ ΑΒΓ κύκλου, καὶ εἰλήφθω τὸ κέντρον τὸ Ε, καὶ ἀπὸ τοῦ Ε ἐπὶ τὴν ΑΓ κάθετος ἤχθω ἡ ΕΖ, καὶ ἐπεζεύχθωσαν αἱ ΕΒ, ΕΓ, ΕΔ· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΕΒΔ. καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΕΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει·
Next, let ΔΓΑ not be through the center of the circle ΑΒΓ, and let the center Ε be found, and from Ε let ΕΖ be drawn perpendicular to ΑΓ, and let ΕΒ, ΕΓ, ΕΔ be joined; therefore the angle ΕΒΔ is right.
ἡ ΑΖ ἄρα τῇ ΖΓ ἐστιν ἴση.
And since a straight line ΕΖ through the center cuts a straight line ΑΓ not through the center at right angles, it also bisects it; therefore ΑΖ is equal to ΖΓ.
καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ζ σημεῖον, πρόσκειται δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΖΕ·
And since the straight line ΑΓ has been bisected at the point Ζ, and ΓΔ is added to it, therefore the rectangle contained by ΑΔ, ΔΓ together with the square on ΖΓ is equal to the square on ΖΔ.
τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τῶν ἀπὸ τῶν ΓΖ, ΖΕ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΔ, ΖΕ. τοῖς δὲ ἀπὸ τῶν ΓΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΓ·
Let the square on ΖΕ be added to each; therefore the rectangle contained by ΑΔ, ΔΓ together with the squares on ΓΖ, ΖΕ is equal to the squares on ΖΔ, ΖΕ.
ὀρθὴ γὰρ ἡ ὑπὸ ΕΖΓ· τοῖς δὲ ἀπὸ τῶν ΔΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΔ· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΔ. ἴση δὲ ἡ ΕΓ τῇ ΕΒ·
But the square on ΕΓ is equal to the squares on ΓΖ, ΖΕ (for the angle ΕΖΓ is right), and the square on ΕΔ is equal to the squares on ΔΖ, ΖΕ; therefore the rectangle contained by ΑΔ, ΔΓ together with the square on ΕΓ is equal to the square on ΕΔ.
τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΔ. τῷ δὲ ἀπὸ τῆς ΕΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΕΒ, ΒΔ·
And ΕΓ is equal to ΕΒ; therefore the rectangle contained by ΑΔ, ΔΓ together with the square on ΕΒ is equal to the square on ΕΔ.
ὀρθὴ γὰρ ἡ ὑπὸ ΕΒΔ γωνία· τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΕΒ, ΒΔ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΕΒ·
But the squares on ΕΒ, ΒΔ are equal to the square on ΕΔ, for the angle ΕΒΔ is right; therefore the rectangle contained by ΑΔ, ΔΓ together with the square on ΕΒ is equal to the squares on ΕΒ, ΒΔ.
λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ.
ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπʼ αὐτοῦ πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφαπτομένης τετραγώνῳ· ὅπερ ἔδει δεῖξαι.
Let the square on ΕΒ be subtracted from each; therefore the remaining rectangle contained by ΑΔ, ΔΓ is equal to the square on ΔΒ. Therefore, if any point is taken outside a circle, and from it two straight lines fall on the circle, and one of them cuts the circle, and the other touches it, the rectangle contained by the whole of the cutting line and the part of it cut off outside between the point and the convex circumference will be equal to the square on the tangent; which was to be proved.