§11## ιά.
## ια΄.
Εἴ κα ἐν κύκλῳ δύο εὐθεῖαι τέμνωντι ἀλλάλας ποτʼ ὀρθὰς μὴ διὰ τοῦ κέντρου οὖσαι, τὰ ἀπὸ τῶν τμαμάτων τῶν εὐθειῶν τετράγωνα τῷ ἀπὸ τᾶς διαμέτρου τοῦ κύκλου ἴσα ἐντί.
If in a circle two straight lines not passing through the center cut each other at right angles, the squares on the segments of the straight lines [together] are equal to the square on the diameter of the circle.
Ἔστω γὰρ κύκλος ὁ ΑΒΓ, καὶ δύο εὐθεῖαι αἱ ΑΒ, Γ△ τετμάσθων ποτʼ ὀρθὰς κατὰ τὸ Ε σαμεῖον· φαμὶ δή, τὰ ἀπὸ τῶν τμαμάτων τῶν ΑΕ, ΕΒ, ΓΕ, Ε△ τετράγωνα τῷ ἀπὸ τᾶς διαμέτρου τοῦ κύκλου ἴσα ἐστίν.
For let there be a circle ABΓ, and let two straight lines AB, ΓΔ cut each other at right angles at the point E; I say indeed that the squares on the segments AE, EB, ΓE, EΔ [together] are equal to the square on the diameter of the circle.
Ἄχθω γὰρ διάμετρος τοῦ induoκύκλου ἁ ΑΖ καὶ ἐπεζεύχθων αἱ ΑΓ, Α△, ΓΖ, △Β εὐθεῖαι.
For let a diameter of the circle AZ be drawn, and let the straight lines AΓ, AΔ, ΓZ, ΔB be joined.
Ἐπεὶ οὖν ἐν δυσὶ τριγώνοις τοῖς Α△Ε, ΑΖΓ γωνίαι αἱ ὑπὸ ΑΕ△, Α△Ε, καὶ ΑΓΖ, ΑΖΓ ἴσαι ἀλλάλαις ἐντὶ ἑκατέρα ἑκατέρᾳ, λοιπαὶ ἄρα γωνίαι αἱ ὑπὸ ΓΑΖ, △ΑΕ ἐσσοῦνται ἀλλάλαις ἴσαι·
Since, then, in the two triangles AΔE, AZΓ the angles AEΔ, AΔE and AΓZ, AZΓ are equal to one another, each to each, therefore the remaining angles ΓAZ, ΔAE will be equal to one another.
περιφέρειαι ἄρα αἱ ΓΖ, △Β ἴσαι ἀλλάλαις ἐντί, καὶ αἱ ταύτας ὑποτείνουσαι εὐθεῖαι αἱ ΓΖ, △Β ἔστι δὲ καὶ τὰ ἀπὸ τῶν △Ε, ΕΒ τῷ ἀπὸ τᾶς △Β, τουτέστι τῷ ἀπὸ τᾶς ΓΖ, ἴσον, καὶ τὰ ἀπὸ τῶν ΑΕ, ΕΓ τῷ ἀπὸ τᾶς ΓΑ, καὶ τὰ ἀπὸ τῶν ΓΖ, ΓΑ τῷ ἀπὸ τᾶς ΖΑ, τουτέστι τῷ ἀπὸ τᾶς διαμέτρου, ἴσα·
Therefore, the arcs ΓZ, ΔB are equal to one another, and also the straight lines ΓZ, ΔB subtending them. And the squares on ΔE, EB [together] are equal to the square on ΔB, that is, to the square on ΓZ, and the squares on AE, EΓ [together] are equal to the square on ΓA, and the squares on ΓZ, ΓA [together] are equal to the square on ZA, that is, to the square on the diameter.
ἐσσοῦνται ἄρα τὰ ἀπὸ τῶν τμαμάτων τῶν ΑΕ, ΕΒ, ΓΕ, Ε△ τετράγωνα τῷ ἀπὸ τᾶς διαμέτρου ἴσα δέδεικται οὖν τὸ προτεθέν.
Therefore, the squares on the segments AE, EB, ΓE, EΔ [together] will be equal to the square on the diameter. Therefore, what was proposed has been proven.
§12## ιβ΄.
## ιβ΄.
Εἴ κα ἐκ σαμείου ἐκτὸς ἁμικυκλίου δύο εὐθεῖαι ἀχθέωντι ἐπιψαύουσαι αὐτοῦ, ἀχθέωντι δὲ ἐκ τῶν σαμείων ἁφᾶς δύο εὐθεῖαι ποτὶ τὰ ἀπεναντίον πέρατα τᾶς διαμέτρου τέμνουσαι ἀλλάλας, ἁ ἐκ τοῦ ἐκτὸς σαμείου ποτὶ τὸ σαμεῖον τομᾶς τῶν δύο εὐθειῶν ἀχθεῖσα καὶ ἐκβληθεῖσα ποτὶ τὰν διάμετρον ἐσσεῖται ταύτᾳ ποτʼ ὀρθάς.
If from a point outside a semicircle two straight lines be drawn tangent to it, and from the points of contact two straight lines be drawn to the opposite extremities of the diameter cutting each other, the line drawn from the outside point to the point of intersection of the two straight lines and produced to the diameter will be at right angles to it.
Ἔστω ἁμικύκλιον τὸ ΑΒ, σαμεῖον δὲ τι ἐκτὸς αὐτοῦ τὸ Γ, καὶ ἐκ τοῦ Γ ἄχθων δύο εὐθεῖαι αἱ Γ△, ΓΕ ἐπίψαύουσαι αὐτοῦ κατὰ τὰ △, Ε σαμεῖα, ἐπεζεύχθων δὲ ἐκ τῶν σαμείων ἁφᾶς ποτὶ τὰ ἀπεναντίον πέρατα τᾶς διαμέτρου τὰ Α, Β εὐθεῖαι αἱ ΕΑ, △Β τέμνουσαι ἀλλάλας κατὰ τὸ Ζ, καὶ ἀχθεῖσα ἁ ΓΖ ἐκβεβλήσθω ἐπὶ τὸ Η σαμεῖον φαμὶ δή, εὐθεῖα ἁ ΓΗ διαμέτρῳ τᾷ ΑΒ ἐσσεῖται ποτʼ ὀρθάς.
Let there be a semicircle AB, and let some point Γ be outside of it, and from Γ let two straight lines ΓΔ, ΓE be drawn tangent to it at the points Δ, E, and let straight lines EA, ΔB be joined from the points of contact to the opposite extremities of the diameter, namely A, B, cutting each other at the point Z, and let the drawn line ΓZ be produced to the point H; I say indeed that the straight line ΓH will be at right angles to the diameter AB.
Ἐπεζεύχθων γὰρ αἱ Α△, ΕΒ. Ἐπεὶ οὖν τριγώνου τοῦ △ΑΒ γωνία ἁ ὑπὸ Α△Β ὀρθά ἐστιν, λοιπαὶ ἄρα γωνίαι αἱ ὑπὸ △ΑΒ, △ΒΑ μιᾷ ὀρθᾷ ἴσαι ἐντί ἔστι δὲ καὶ γωνία ἁ ὑπὸ ΑΕΒ μιᾷ ὀρθᾷ ἴσα κοινὰ ποτικείσθω ἁ ὑπὸ ΖΒΕ συναμφότερος ἄρα ἁ ὑπὸ △ΑΒ, ΑΒΕ συναμφοτέρῳ τᾷ ὑπὸ ΖΒΕ, ΖΕΒ, τουτέστιν ἐξωτερικᾷ γωνίᾳ τᾷ ὑπὸ △ΖΕ τριγώνου τοῦ ΖΒΕ ἐστὶν ἴσα.
For let AΔ, EB be joined. Since, then, the angle AΔB of the triangle ΔAB is a right angle, therefore the remaining angles ΔAB, ΔBA are equal to one right angle. And the angle AEB is also equal to one right angle; let ΖBE be added as common. Therefore, the combined angle ΔAB, ABE is equal to the combined angle ΖBE, ΖEB, that is, to the exterior angle ΔΖE of the triangle ΖBE.
Καὶ ἐπεὶ εὐθεῖα ἁ Γ△ ἐπιψαύουσα τοῦ κύκλου ἐστίν, διᾶκται δὲ ἀπὸ τοῦ σαμείου ἁφᾶς τοῦ κλου ἁ △Β τέμνουσα τὸν κύκλον, ἐσσεῖται γωνία ἁ ὑπὸ Γ△Β γωνίᾳ τᾷ ὑπὸ △ΑΒ ἴσα διὰ τὰ αὐτὰ δὴ καὶ γωνία ἁ ὑπὸ ΓΕΖ τᾷ ὑπὸ ΕΒΑ ἐστὶν ἴσα καὶ συναμφότερος ἄρα γωνία ἁ ὑπὸ ΓΕΖ, Γ△Ζ τᾷ ὑπὸ △ΖΕ ἐστὶν ἴσα καὶ δέδεικται παῤ ἡμῶν ἐν τοῖς Περὶ τετραπλεύρων ὅτι εἴ κα μεταξὺ δύο ἰσᾶν εὐθειᾶν τεμνομενᾶν, οἷον τᾶν Γ△, ΓΕ, δύο εὐθεῖαι ἀχθέωντι τεμνόμεναί, οἷον αἱ △Ζ, ΕΖ, γωνία δὲ ἁ ὑπὸ τούτων περιεχομὲνα, ὡς ἁ ποτὶ τῷ Ζ, συναμφοτέρῳ τᾷ ὑπὸ τῶν δύο τεμνομενᾶν εὐθειᾶν περιεχομένᾳ, ὡς αἱ ποτὶ τοῖς Ε, △ σαμείοις, ἴσα ἐστίν, ἁ ἐπιζευγνυμένα ἐκ τοῦ σαμείου καθʼ ὃ αἱ δύο εὐθεῖαι συμβάλλοντι ἐπὶ τὸ σαμεῖον καθʼ ὃ αὗται τὲμνοντι ἀλλάλας, ὡς ἁ ΓΖ εὐθεῖα, ἑκατέρᾳ τᾶν τεμνομενᾶν εὐθειᾶν, ὡς αἱ Γ△, ΓΕ, ἐστὶν ἴσα·
And since the straight line ΓΔ is tangent to the circle, and from the point of contact there is drawn ΔB cutting the circle, the angle ΓΔB will be equal to the angle ΔAB. For the very same reasons indeed, the angle ΓEΖ is also equal to EBA. Therefore, the combined angle ΓEΖ, ΓΔΖ is also equal to ΔΖE. And it has been proven by us in "On Quadrilaterals" that if between two equal intersecting straight lines, such as ΓΔ, ΓE, two straight lines be drawn intersecting, such as ΔΖ, EΖ, and the angle contained by them, such as that at Ζ, is equal to the combined angle contained by the two intersecting straight lines, such as those at the points E, Δ, then the line joining from the point where the two straight lines meet to the point where they cut each other, such as the straight line ΓΖ, is equal to each of the intersecting straight lines, such as ΓΔ, ΓE.
ἁ ΓΖ εὐθεῖα ἄρα τᾷ Γ△ ἐστὶν ἴσα καὶ γωνία ἁ ὑπὸ ΓΖ△ γωνίᾳ τᾷ ὑπὸ Γ△Ζ, τουτέστι τᾷ ὑπὸ △ΑΗ·
Therefore, the straight line ΓΖ is equal to ΓΔ. And the angle ΓΖΔ is also equal to the angle ΓΔΖ, that is, to ΔAH.
γωνίαι δὲ αἱ ὑπὸ ΓΖ△, △ΖΗ δυσὶν ὀρθαῖς ἴσαι ἐντί·
And the angles ΓΖΔ, ΔΖH are equal to two right angles.
συναμφότερος ἄρα γωνία ἁ ὑπὸ △ΑΗ, △ΖΗ δυσὶν ὀρθαῖς ἴσα ἐστίν·
Therefore, the combined angle ΔAH, ΔΖH is equal to two right angles.
λοιπαὶ ἄρα γωνίαι τετραπλεύρου τοῦ Α△ΖΗ αἱ ὑπὸ Α△Ζ, ΑΗΖ δυσὶν ὀρθαῖς ἴσαι ἐντί·
Therefore, the remaining angles of the quadrilateral AΔΖH, namely AΔΖ, AHΖ, are equal to two right angles.
ἔστι δὲ γωνία ἁ ὑπὸ Α△Β μιᾷ ὀρθᾷ ἴσα γωνία ἄρα ἁ ὑπὸ ΑΗΓ μιᾷ ὀρθᾷ ἴσα ἐστίν·
And the angle AΔB is equal to one right angle; therefore the angle AHΓ is equal to one right angle.
ἔστιν ἄρα εὐθεῖα ἁ △Η διαμέτρῳ τᾷ ΑΒ ποτʼ ὀρθάς·
Therefore, the straight line ΔH is at right angles to the diameter AB.
δέδεικται οὖν τὸ προτεθέν.
Therefore, what was proposed has been proven.