§4## δ΄.
## δ΄.
Εἴ κα ἐν ἁμικυκλίῳ σαμεῖόν· τι ἐπὶ τᾶς διαμέτρου ᾖ, γραφέωντι δὲ ἀπὸ τῶν τμαμάτων τᾶς διαμέτρου δύο ἁμικύκλια ἐντός, ἀναστακῇ δὲ ἀπὸ τοῦ λαφθέντος σαμείου εὐθεῖα ποτὶ τᾷ περιφερείᾳ τᾷ διαμέτρῳ ποτʼ ὀρθάς, σχῆμα τὸ ὑπὸ τῶν τριῶν περιφερειῶν περιεχόμενον ἴσον ἐστὶ κύκλῳ, οὗ διάμετρος ἁ ἀναστακεῖσα κάθετος.
If in a semicircle there be any point on the diameter, and two semicircles are described internally from the segments of the diameter, and a straight line is erected from the taken point perpendicular to the diameter and meeting the circumference, the figure bounded by the three circumferences is equal to the circle whose diameter is the erected perpendicular.
Ἔστω ἁμικύκλιον τὸ ΑΒΓ καὶ σαμεῖόν τι ἐπὶ διαμέτρου τᾶς ΑΓ τὸ △, καὶ ἀπὸ διαμέτρων τῶν Γ△, △Α ἁμικύκλια ἀναγεγράφθων ἐντός, ἀπὸ δὲ τοῦ △ σαμείου ἀνεστακέτω ποτʼ ὀρθὰς τᾷ ΑΓ ἁ △Β φαμὶ δή, σχῆμα τὸ ὑπὸ τῶν τριῶν περιφερειῶν περιεχόμενον, τουτέστι τοῦ μείζονος ἁμικυκλίου καὶ τῶν δύο ἀναγραφέντων ἐντός, ὅπερ Et ἄρβηλος καλείσθω, κύκλῳ, οὗ διάμετρος ἁ △Β, ἴσον ἐστίν.
Let there be a semicircle ABΓ, and any point Δ on the diameter AΓ, and let semicircles be described internally from the diameters ΓΔ, ΔA, and from the point Δ let ΔB be erected perpendicular to AΓ. I say that the figure bounded by the three circumferences, that is, by the greater semicircle and the two described internally (which let be called "arbelos"), is equal to the circle whose diameter is ΔB.
Ἐπεὶ γὰρ εὐθεῖαι αἱ △Α, △Β, △Γ ἑξῆς ἀνάλογόν ἐντι, ἐσσεῖται τὸ ὑπὸ τῶν Α△, △Γ τῷ ἀπὸ τᾶς Β△ ἴσον·
For since the straight lines ΔA, ΔB, ΔΓ are in continued proportion, the rectangle contained by AΔ, ΔΓ will be equal to the square on BΔ.
κοινὸν ποτικείσθω τὸ ὑπὸ τῶν Α△, △Γ καὶ τὰ ἀπὸ τῶν Α△, △Γ·
Let the common rectangle contained by AΔ, ΔΓ and the squares on AΔ, ΔΓ be added.
τὸ ἄρα ἀπὸ τᾶς ὅλας τετράγωνον, τουτέστι τὸ ἀπὸ τᾶς ΑΓ, τοῖς ἀπὸ τῶν τμαμάτων τῶν ἀπὸ τῶν Α△, △Γ τετραγώνοις καὶ τῷ δὶς τοῦ ἀπὸ τᾶς Β△ ἐστὶν ἴσον.
Therefore, the square on the whole, that is, on AΓ, is equal to the squares on the segments AΔ, ΔΓ and twice the square on BΔ.
Καὶ ἐπεὶ οἱ κύκλοι πρὸς ἀλλάλους ὡς τὰ ἀπὸ τᾶν διαμέτρων τετράγωνά ἐντι, ἐσσεῖται δὴ κύκλος, οὗ διάμετρος ἁ ΑΓ, δυσὶ κύκλοις, ὧν διάμετρος ἁ △Β, καὶ δυσὶ κύκλοις, ὧν διάμετροι αἱ Α△, △Γ, ἴσος, τουτέστιν ἁμικύκλιον τὸ ΑΓ ἴσον κύκλῳ, οὗ διάμετρος ἁ △Β, καὶ δυσὶν ἁμικυκλίοις, ὧν διάμετροι αἱ Α△, △Γ·
And since circles are to one another as the squares on their diameters, the circle whose diameter is AΓ will be equal to two circles whose diameter is ΔB, and two circles whose diameters are AΔ, ΔΓ; that is, the semicircle AΓ is equal to the circle whose diameter is ΔB, and two semicircles whose diameters are AΔ, ΔΓ.
κοινὸν ἀφαιρήσθω ἁμικύκλια τὰ Α△, △Γ·
Let the common semicircles AΔ, ΔΓ be subtracted.
λοιπὸν ἄρα χωρίον τὸ περιεχόμενον ὑπὸ περιφερειῶν τᾶν ΑΓ, Α△, △Γ, ὅπερ ἄρβηλος καλεῖται, κύκλῳ, οὗ διάμετρος ἁ △Β, ἐστὶν ἴσον·
Therefore, the remaining area bounded by the circumferences AΓ, AΔ, ΔΓ, which is called arbelos, is equal to the circle whose diameter is ΔB.
δέδεικται οὖν τὸ προτεθέν.
Therefore, what was proposed has been proven.
§5## ε΄.
## ε΄.
Εἴ κα ἐν ἁμικυκλίῳ σαμεῖόν τι ἐπὶ τᾶς διαμέτρου ᾖ, καὶ γραφέωντι ἀπὸ τῶν τμαμάτων τᾶς διαμέτρου δύο ἁμικύκλια ἐντός, ἀναστακῇ δὲ ἀπὸ τοῦ σαμείου εὐθεῖα τᾷ διαμέτρῳ ποτ᾿ ὀρθάς, καὶ δύο κύκλοι γραφέωντι ἐπʼ ἀμφότερα τᾶς ἀνεστακούσας ἐπιψαύοντες αὐτᾶς καὶ τῶν ἁμικυκλίων, οἱ γραφέντες κύκλοι ἐσσοῦνται ἀλλάλοις ἴσοι.
If in a semicircle there be any point on the diameter, and two semicircles are described internally from the segments of the diameter, and a straight line is erected from the point perpendicular to the diameter, and two circles are described on both sides of the erected line touching it and the semicircles, the described circles will be equal to each other.
Ἔστω ἁμικύκλιον, οὗ διάμετρος ἁ ΑΒ, σαμεῖον δέ τι ἐπʼ αὐτᾶς τὸ Γ·
Let there be a semicircle whose diameter is AB, and any point Γ on it.
ἀναγεγράφθω δὲ ἀπὸ τμαμάτων τῶν ΑΓ, ΓΒ ἁμικύκλια ἐντός, καὶ ἀπὸ τοῦ Γ σαμείου ἀνεστακέτω ποτʼ ὀρθὰς διαμέτρω τᾷ ΑΒ ἁ Γ△, γεγράφθων δὲ δύο κύκλοι ἐπʼ ἀμφότερα τᾶς ἀνεστακούσας εὐθείας ἐπιψαύοντες τᾶς τε ἀνεστακούσας καὶ τῶν ἁμικυκλίων·
And let semicircles be described internally from the segments AΓ, ΓB, and from the point Γ let ΓΔ be erected perpendicular to the diameter AB, and let two circles be described on both sides of the erected straight line touching both the erected line and the semicircles.
φαμὶ δή, οἱ γραφέντες κύκλοι ἴσοι ἀλλάλοις ἐντί.
I say that the described circles are equal to each other.
Ἔστω γὰρ πρότερον κύκλος ὁ ἐπιψαύων τᾶς Γ△ κατὰ τὸ Ε σαμεῖον καὶ ἁμικυκλίου μὲν τοῦ ΑΓ κατὰ τὸ H, ἁμικυκλίου δὲ τοῦ ΑΒ κατὰ τὸ Ζ, ἄχθω δὲ διάμετρος τοῦ κύκλου ἁ ΘΕ·
For let there first be the circle touching ΓΔ at the point E, and the semicircle AΓ at H, and the semicircle AB at Z, and let the diameter of the circle be ΘE.
ἐπιζευχθεῖσαι δὴ αἱ ΑΘ, ΘΖ εὐθεῖαι ἐσσοῦνται ἀλλάλαις ἐπʼ εὐθείας, ἐκβληθεῖσαι δὲ αἱ ΑΖ, ΓΕ εὐθεῖαι συμβαλέτωσαν κατὰ τὸ △ σαμεῖον ὁμοίως δὴ ἐπιζευχθεῖσαι αἱ ΖΕ, ΕΒ ἐσσοῦνται ἀλλάλαις ἐπʼ εὐθείας, καὶ αἱ ΘΗ, ΗΓ, καὶ αἱ ΕΗ. ΗΑ, ἐκβεβλήσθω δὲ ἁ ΑΕ ἐπὶ τὸ Ι σαμεῖον, ἄχθω δὲ ἁ ΒΙ εὐθεῖα καὶ ἁ Ζ△.
Therefore, the joined straight lines AΘ, ΘZ will be in a straight line with each other, and let the produced straight lines AZ, ΓE meet at the point Δ. Similarly, the joined ZE, EB will be in a straight line with each other, and also ΘH, HΓ, and EH, HA. And let AE be produced to the point I, and let the straight line BI and ZΔ be drawn.
Ἐπεὶ οὖν αἱ Α△, ΑΒ εὐθεῖαί ἐντι καὶ ἀπὸ τοῦ △ σαμείου τᾷ ΑΒ ἆκται ποτʼ ὀρθὰς ἁ △Γ, καὶ ἀπὸ τοῦ Β ποτʼ ὀρθὰς τᾷ △Α ἁ ΒΖ τέμνουσα τὰν △Γ Bl, κατὰ τὸ Ε, εὐθεῖα δὲ ἁ ΑΕΙ ποτʼ ὀρθὰς τᾷ ΒΙ ἐστίν, ἐσσοῦνται ἄρα εὐθεῖαι αἱ ΒΙ, Ι△ ἀλλάλαις ἐπʼ εὐθείας, ὡς παῤ ἡμῶν ἐν τοῖς Περὶ ὀρθογωνίων τριγώνων δέδεικται Καὶ ἐπεὶ εὐθεῖα ἁ Β△ παρὰ τὰν ΓΗ ἐστίν, τὸν αὐτὸν λόγον ἔχει ἁ Α△ ποτὶ τὰν △Θ, ὃν ἔχει ἁ ΑΓ ποτὶ τὰν ΘΕ, τουτέστιν ἁ ΑΒ ποτὶ τὰν ΒΓ τὸ ἄρα ὑπὸ τᾶν ΑΓ, ΓΒ τῷ ὑπὸ τᾶν ΑΒ, ΘΕ ἐστὶν ἴσον·
Since, then, AΔ, AB are straight lines, and from the point Δ, ΔΓ has been drawn perpendicular to AB, and from B, BZ has been drawn perpendicular to ΔA, cutting ΔΓ (Bl) at E, and the straight line AEI is perpendicular to BI, therefore the straight lines BI, IΔ will be in a straight line with each other, as has been proven by us in our work "On Right-Angled Triangles." And since the straight line BΔ is parallel to ΓH, AΔ has the same ratio to ΔΘ as AΓ has to ΘE, that is, as AB has to BΓ. Therefore, the rectangle contained by AΓ, ΓB is equal to the rectangle contained by AB, ΘE.
ὅμοίως δὴ δείξομες ὅτι ἐν κύκλῳ τῷ ΛΜΝ τὸ ὑπὸ τᾶν ΑΓ, ΓΒ τῷ ὑπὸ ΑΒ καὶ τᾶς διαμέτρου τοῦ ΛΜΝ κύκλου ἴσον ἐστίν·
In like manner we shall show that in the circle ΛΜΝ, the rectangle contained by AΓ, ΓB is equal to the rectangle contained by AB and the diameter of the circle ΛΜΝ.
αἱ διάμετροι ἄρα κύκλων τῶν ἴσαι ἐντί, τουτέστιν οἱ δύο κύκλοι ἴσοι ἐντί·
Therefore, the diameters of the circles are equal, that is, the two circles are equal.
δέδεικται οὖν τὸ προτεθέν.
Therefore, what was proposed has been proven.