§1## α΄.
## α΄.
Εἴ κα ᾗ δύο κύκλοι ἐπιψαύοντες ἀλλάλων ἐντός, διάμετροι δὲ αὐτῶν παράλληλοι, ἐπιζευχθεῖσαι αἱ ἀπὸ τοῦ σαμείου ἁφῆς καὶ τῶν περάτων τῶν διαμέτρων δύο εὐθεῖαι ἐσσοῦνται ἀλλάλαις ἐπʼ εὐθείας.
If two circles touch each other internally, and their diameters are parallel, the two straight lines joined from the point of contact to the extremities of the diameters will be in a straight line with each other.
Ἔστωσαν δύο κύκλοι, ὧν κέντρα τὰ Ζ, Η, ἐπιψαύοντες ἀλλάλων κατὰ τὸ Ε σαμεῖον, διάμετρος δὲ ἁ ΑΒ παρὰ διάμετρον τὰν Γ△·
Let there be two circles, whose centers are Z, H, touching each other at the point E, and let the diameter AB be parallel to the diameter ΓΔ.
φαμὶ δή, ἐπιζευχθεῖσαι αἱ Ε△, △Β εὐθεῖαι ἐσσοῦνται ἀλλάλαις ἐπʼ εὐθείας.
I say that the straight lines EΔ, ΔB, being joined, will be in a straight line with each other.
Ἐπεζεύχθω γὰρ ἁ ΖΗ καὶ ἐκβεβλήσθω ποτὶ τὸ Ε, ἄχθω δὲ ἁ △Θ παρὰ τὰν ΖΗ.
Ἐπεὶ οὖν εὐθεῖαι αἱ ΖΒ, ΖΕ ἴσαι ἐντὶ καὶ ἁ H△ τᾷ ΖΘ, κοινὰ ἀφαιρήσθω ἁ ΖΘ, τουτέστιν ἁ ΗΕ·
For let ZH be joined and produced to E, and let ΔΘ be drawn parallel to ZH. Since, then, the straight lines ZB, ZE are equal, and HΔ is equal to ZΘ, let the common part ZΘ, which is HE, be subtracted.
λοιπαὶ ἄρα εὐθεῖαι αἱ Θ△, ΘΒ ἴσαι ἀλλαλαις ἐντί·
Therefore, the remaining straight lines ΘΔ, ΘB are equal to each other.
γωνία ἄρα ἁ ὑπὸ Θ△Β γωνίᾳ τᾷ ὑπὸ ΘΒ△, τουτέστιν τᾷ ὑπὸ H△Ε, ἐστὶν ἴσα·
Therefore, the angle ΘΔB is equal to the angle ΘBΔ, that is, to the angle HΔE.
κοινὰ ποτικείσθω γωνία ἁ ὑπὸ H△Β συναμφότερος ἄρα γωνία ἁ ὑπὸ H△Β, △ΒΖ συναμφοτέρῳ τᾷ ὑπὸ Η△Β. Ε△Η ἐστὶν ἴσα·
Let the common angle HΔB be added. Therefore, the sum of the angles HΔB, ΔBZ is equal to the sum of the angles HΔB, EΔH.
ἔστι δὲ συναμφότερος ἁ ὑπὸ Η△Β, △ΒΖ δυσὶν ὀρθαῖς ἴσα·
But the sum of the angles HΔB, ΔBZ is equal to two right angles.
συναμφότερος ἄρα γωνία ἁ ὑπὸ Η△Β, Ε△Η δυσὶν ὀρθαῖς ἐστιν ἴσα·
Therefore, the sum of the angles HΔB, EΔH is also equal to two right angles.
ἐπʼ εὐθείας ἄρα ἐντὶ εὐθεῖαι αἱ Ε△, △Β·
Therefore, the straight lines EΔ, ΔB are in a straight line.
δέδεικται οὖν τὸ προτεθέν.
Therefore, what was proposed has been proven.
§2## β΄.
## β΄.
Ἔστω ἁμικύκλιον τὸ ΑΒΓ καὶ δύο εὐθεῖαι ἐπιψαύουσαι αὐτοῦ αἱ △Β, △Γ, ἁ δὲ ΒΕ ἄχθω ποτʼ ὀρθὰς τᾷ ΑΓ, ἐπεζεύχθω δὲ ἁ Α△·
Let there be a semicircle ABΓ, and two straight lines ΔB, ΔΓ touching it, and let BE be drawn perpendicular to AΓ, and let AΔ be joined.
φαμὶ δὴ τὰν ΒΖ ἴσαν εἶμεν τᾷ ΖΕ.
Ἐπεζεύχθω γὰρ ἁ ΑΒ καὶ ἐκβληθεῖσαι αἱ ΑΒ, Γ△ συμπιπτέτωσαν κατὰ τὸ Η σαμεῖον καὶ ἄχθω ἁ ΒΓ.
Ἐπεὶ οὖν γωνία ἁ ὑπὸ ὀρθά ἐστιν, ἐσσεῖται καὶ γωνία ἁ ὑπὸ ΓΒΗ ὀρθά ἔστι δὲ καὶ εὐθεῖα ἁ Β△ τᾷ △Γ ἴσα·
I say that BZ is equal to ZE. For let AB be joined, and let AB, ΓΔ, being produced, meet at the point H, and let BΓ be drawn. Since, then, the angle ABΓ is a right angle, the angle ΓBH will also be a right angle. And the straight line BΔ is equal to ΔΓ.
ἐσσεῖται ἄρα καὶ εὐθεῖα ἁ △Η τᾷ △Β, τουτέστι τᾷ △Γ ἴσα.
Therefore, the straight line ΔH will also be equal to ΔB, that is, to ΔΓ.
Καὶ ἐπεὶ ἁ ΒΕ παρὰ τὰν ΗΓ ἐστίν, ἐσσεῖται ἄρα καὶ ἁ ΒΖ τᾷ ΖΕ ἴσα·
And since BE is parallel to HΓ, therefore BZ will also be equal to ZE.
δέδεικται οὖν τὸ προτεθέν.
Therefore, what was proposed has been proven.
§3## γ΄.
## γ΄.
Ἔστω τμᾶμα κύκλου τὸ ΑΓ καὶ ἀπὸ σαμείου τινος Β τᾶς περιφερείας ἄχθω τᾷ ΑΓ ποτ᾿ ὀρθὰς ἁ Β△, λελάφθω δὲ εὐθεῖα ἁ △Ε εὐθείᾳ τᾷ △Α ἴσα καὶ περιφέρεια ἁ ΒΖ τᾷ ΑΒ·
Let there be a segment of a circle AΓ, and from any point B on its circumference let BΔ be drawn perpendicular to AΓ, and let the straight line ΔE be taken equal to the straight line ΔA, and the arc BZ equal to AB.
φαμὶ δή, ἐπιζευχθεῖσα ἁ ΓΖ εὐθεῖα τᾷ ΓΕ ἐστὶν ἴσα.
I say that the straight line ΓZ, being joined, is equal to ΓE.
Ἐπεζεύχθωσαν γὰρ αἱ ΑΒ, ΒΖ, ΖΕ, ΕΒ εὐθεῖαι·
For let the straight lines AB, BZ, ZE, EB be joined.
καὶ ἐπεὶ ἁ Α△ τᾷ △Ε ἐστὶν ἴσα, κοινὰ δὲ ἁ Β△, δύο δὴ αἱ Α△, △Β δυσὶ ταῖς Ε△, △Β ἑκατέρα ἑκατέρᾳ ἴσαι ἐντί·
And since AΔ is equal to ΔE, and BΔ is common, the two straight lines AΔ, ΔB are equal to the two straight lines EΔ, ΔB respectively.
ἔστι δὲ γωνία ἁ ὑπὸ Α△Β γωνίᾳ τᾷ ὑπὸ Ε△Β ἴσα βάσις ἄρα ἁ ΕΒ βάσει τᾷ ΑΒ, τουτέστι τᾷ ΒΖ, ἐστὶν ἴσα γωνία ἄρα ἁ ὑπὸ ΒΕΖ γωνίᾳ τᾷ ὑπὸ ΒΖΕ ἐστὶν ἴσα.
And the angle AΔB is equal to the angle EΔB. Therefore, the base EB is equal to the base AB, that is, to BZ. Therefore, the angle BEZ is equal to the angle BZE.
Καὶ ἐπεὶ τετράπλευρον τὸ ΑΒΖΓ ἐν κύκλῳ ἐστίν, γωνίαι αἱ ἀπεναντίον αἱ ὑπὸ ΓΖΒ, ΓΑΒ, τουτέστιν αἱ ὑπὸ ΓΖΒ, ΒΕΑ, δυσὶν ὀρθαῖς ἴσαι ἐντί.
And since the quadrilateral ABZΓ is in a circle, the opposite angles ΓZB and ΓAB, that is, ΓZB and BEA, are equal to two right angles.
Ἔστι δὲ καὶ συναμφότερος ἁ ὑπὸ ΓΕΒ, ΒΕΑ δυσὶν ὀρθαῖς ἴσα κοινὰ ἀφαιρήσθω ἁ ὑπὸ ΒΕΑ γωνία ἄρα ἁ ὑπὸ ΓΖΒ γωνίᾳ τᾷ ὑπὸ ΓΕΒ ἐστὶν ἴσα κοινὰ ἀφαιρήσθω ἁ ὑπὸ ΒΖΕ, τουτέστιν ἁ ὑπὸ ΒΕΖ·
But the sum of the angles ΓEB and BEA is also equal to two right angles. Let the common angle BEA be subtracted. Therefore, the angle ΓZB is equal to the angle ΓEB. Let the common angle BZE, which is BEZ, be subtracted.
λοιπαὶ ἄρα αἱ ποτὶ τᾷ βάσει τᾷ ΕΖ τριγώνου τοῦ ΕΓΖ· γωνίαι αἱ ὑπὸ ΓΖΕ, ΖΕΓ ἴσαι ἀλλάλαις ἐντί·
Therefore, the remaining angles at the base EZ of the triangle EΓZ, namely the angles ΓZE and ZEΓ, are equal to each other.
πλευρὰ ἄρα ἁ ΖΓ πλευρᾷ τᾷ ΕΓ ἐστὶν ἴσα δέδεικται οὖν τὸ προτεθέν.
Therefore, the side ZΓ is equal to the side EΓ. Therefore, what was proposed has been proven.