§3.prop.23ἐπὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων ὅμοια καὶ ἄνισα οὐ συσταθήσεται ἐπὶ τὰ αὐτὰ μέρη.
On the same straight line there shall not be constructed two similar and unequal segments of circles on the same side.
εἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ δύο τμήματα κύκλων ὅμοια καὶ ἄνισα συνεστάτω ἐπὶ τὰ αὐτὰ μέρη τὰ ΑΓΒ, ΑΔΒ, καὶ διήχθω ἡ ΑΓΔ, καὶ ἐπεζεύχθωσαν αἱ ΓΒ, ΔΒ.
ἐπεὶ οὖν ὅμοιόν ἐστι τὸ ΑΓΒ τμῆμα τῷ ΑΔΒ τμήματι, ὅμοια δὲ τμήματα κύκλων ἐστὶ τὰ δεχόμενα γωνίας ἴσας, ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΑΔΒ ἡ ἐκτὸς τῇ ἐντός· ὅπερ ἐστὶν ἀδύνατον.
For, if possible, on the same straight line ΑΒ let two similar and unequal segments of circles ΑΓΒ, ΑΔΒ be constructed on the same side, and let ΑΓΔ be drawn through, and let ΓΒ, ΔΒ be joined. Since then the segment ΑΓΒ is similar to the segment ΑΔΒ, and similar segments of circles are those which admit equal angles, the angle ΑΓΒ is therefore equal to the angle ΑΔΒ, the exterior to the interior; which is impossible.
οὐκ ἄρα ἐπὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων ὅμοια καὶ ἄνισα συσταθήσεται ἐπὶ τὰ αὐτὰ μέρη· ὅπερ ἔδει δεῖξαι.
There shall not therefore be constructed on the same straight line two similar and unequal segments of circles on the same side; which was to be proved.
§3.prop.24τὰ ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν.
Similar segments of circles on equal straight lines are equal to one another.
ἔστωσαν γὰρ ἐπὶ ἴσων εὐθειῶν τῶν ΑΒ, ΓΔ ὅμοια τμήματα κύκλων τὰ ΑΕΒ, ΓΖΔ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΕΒ τμῆμα τῷ ΓΖΔ τμήματι.
For let ΑΕΒ, ΓΖΔ be similar segments of circles on equal straight lines ΑΒ, ΓΔ; I say that the segment ΑΕΒ is equal to the segment ΓΖΔ.
Ἐφαρμοζομένου γὰρ τοῦ ΑΕΒ τμήματος ἐπὶ τὸ ΓΖΔ καὶ τιθεμένου τοῦ μὲν Α σημείου ἐπὶ τὸ Γ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΓΔ, ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Δ σημεῖον διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΓΔ· τῆς δὲ ΑΒ ἐπὶ τὴν ΓΔ ἐφαρμοσάσης ἐφαρμόσει καὶ τὸ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ. εἰ γὰρ ἡ ΑΒ εὐθεῖα ἐπὶ τὴν ΓΔ ἐφαρμόσει, τὸ δὲ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ μὴ ἐφαρμόσει, ἤτοι ἐντὸς αὐτοῦ πεσεῖται ἢ ἐκτὸς ἢ παραλλάξει ὡς τὸ ΓΗΔ, καὶ κύκλος κύκλον τέμνει κατὰ πλείονα σημεῖα ἢ δύο· ὅπερ ἐστὶν ἀδύνατον.
For, if the segment ΑΕΒ be applied to ΓΖΔ, and the point Α be placed on Γ and the straight line ΑΒ on ΓΔ, the point Β will also coincide with the point Δ because ΑΒ is equal to ΓΔ; and, ΑΒ coinciding with ΓΔ, the segment ΑΕΒ will also coincide with ΓΖΔ. For if the straight line ΑΒ coincide with ΓΔ, but the segment ΑΕΒ do not coincide with ΓΖΔ, it will either fall within it, or without, or it will deviate as ΓΗΔ, and a circle cuts a circle at more points than two; which is impossible.
οὐκ ἄρα ἐφαρμοζομένης τῆς ΑΒ εὐθείας ἐπὶ τὴν ΓΔ οὐκ ἐφαρμόσει καὶ τὸ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ· ἐφαρμόσει ἄρα, καὶ ἴσον αὐτῷ ἔσται.
Therefore, if the straight line ΑΒ is applied to ΓΔ, the segment ΑΕΒ will not fail to coincide with ΓΖΔ; therefore it will coincide, and will be equal to it.
τὰ ἄρα ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
Therefore similar segments of circles on equal straight lines are equal to one another; which was to be proved.
§3.prop.25κύκλου τμήματος δοθέντος προσαναγράψαι τὸν κύκλον, οὗπέρ ἐστι τμῆμα.
Given a segment of a circle, to describe the complete circle of which it is a segment.
ἔστω τὸ δοθὲν τμῆμα κύκλου τὸ ΑΒΓ· δεῖ δὴ τοῦ ΑΒΓ τμήματος προσαναγράψαι τὸν κύκλον, οὗπέρ ἐστι τμῆμα.
Let the given segment of a circle be ΑΒΓ; it is then required to describe the complete circle of which the segment ΑΒΓ is a segment.
τετμήσθω γὰρ ἡ ΑΓ δίχα κατὰ τὸ Δ, καὶ ἤχθω ἀπὸ τοῦ Δ σημείου τῇ ΑΓ πρὸς ὀρθὰς ἡ ΔΒ, καὶ ἐπεζεύχθω ἡ ΑΒ· ἡ ὑπὸ ΑΒΔ γωνία ἄρα τῆς ὑπὸ ΒΑΔ ἤτοι μείζων ἐστὶν ἢ ἴση ἢ ἐλάττων.
For let ΑΓ be cut in half at Δ, and from the point Δ let ΔΒ be drawn at right angles to ΑΓ, and let ΑΒ be joined; the angle ΑΒΔ is therefore either greater than, or equal to, or less than the angle ΒΑΔ.
ἔστω πρότερον μείζων, καὶ συνεστάτω πρὸς τῇ ΒΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΒΔ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΕ, καὶ διήχθω ἡ ΔΒ ἐπὶ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΕΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία τῇ ὑπὸ ΒΑΕ, ἴση ἄρα ἐστὶ καὶ ἡ ΕΒ εὐθεῖα τῇ ΕΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΓ, κοινὴ δὲ ἡ ΔΕ, δύο δὴ αἱ ΑΔ, ΔΕ δύο ταῖς ΓΔ, ΔΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΑΔΕ γωνίᾳ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση· ὀρθὴ γὰρ ἑκατέρα· βάσις ἄρα ἡ ΑΕ βάσει τῇ ΓΕ ἐστιν ἴση.
Let it first be greater, and on the straight line ΒΑ and at the point Α on it let the angle ΒΑΕ be constructed equal to the angle ΑΒΔ, and let ΔΒ be produced to Ε, and let ΕΓ be joined. Since then the angle ΑΒΕ is equal to the angle ΒΑΕ, the straight line ΕΒ is also equal to ΕΑ. And since ΑΔ is equal to ΔΓ, and ΔΕ is common, the two ΑΔ, ΔΕ are equal to the two ΓΔ, ΔΕ respectively; and the angle ΑΔΕ is equal to the angle ΓΔΕ, for each is right; therefore the base ΑΕ is equal to the base ΓΕ.
ἀλλὰ ἡ ΑΕ τῇ ΒΕ ἐδείχθη ἴση· καὶ ἡ ΒΕ ἄρα τῇ ΓΕ ἐστιν ἴση· αἱ τρεῖς ἄρα αἱ ΑΕ, ΕΒ, ΕΓ ἴσαι ἀλλήλαις εἰσίν·
But ΑΕ was proved equal to ΒΕ; therefore ΒΕ is also equal to ΓΕ; therefore the three ΑΕ, ΕΒ, ΕΓ are equal to one another.
ὁ ἄρα κέντρῳ τῷ Ε διαστήματι δὲ ἑνὶ τῶν ΑΕ, ΕΒ, ΕΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται προσαναγεγραμμένος.
Therefore the circle described with center Ε and with distance one of ΑΕ, ΕΒ, ΕΓ will also pass through the remaining points and will be completed.
κύκλου ἄρα τμήματος δοθέντος προσαναγέγραπται ὁ κύκλος.
Therefore, given a segment of a circle, the circle has been described.
καὶ δῆλον, ὡς τὸ ΑΒΓ τμῆμα ἔλαττόν ἐστιν ἡμικυκλίου διὰ τὸ τὸ Ε κέντρον ἐκτὸς αὐτοῦ τυγχάνειν.
And it is manifest that the segment ΑΒΓ is less than a semicircle because the center Ε happens to be outside it.
ὁμοίως κἂν ᾖ ἡ ὑπὸ ΑΒΔ γωνία ἴση τῇ ὑπὸ ΒΑΔ, τῆς ΑΔ ἴσης γενομένης ἑκατέρᾳ τῶν ΒΔ, ΔΓ αἱ τρεῖς αἱ ΔΑ, ΔΒ, ΔΓ ἴσαι ἀλλήλαις ἔσονται, καὶ ἔσται τὸ Δ κέντρον τοῦ προσαναπεπληρωμένου κύκλου, καὶ δηλαδὴ ἔσται τὸ ΑΒΓ ἡμικύκλιον.
Similarly, even if the angle ΑΒΔ be equal to ΒΑΔ, since ΑΔ becomes equal to each of ΒΔ, ΔΓ, the three ΔΑ, ΔΒ, ΔΓ will be equal to one another, and Δ will be the center of the completed circle, and obviously ΑΒΓ will be a semicircle.
ἐὰν δὲ ἡ ὑπὸ ΑΒΔ ἐλάττων ᾖ τῆς ὑπὸ ΒΑΔ, καὶ συστησώμεθα πρὸς τῇ ΒΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΒΔ γωνίᾳ ἴσην, ἐντὸς τοῦ ΑΒΓ τμήματος πεσεῖται τὸ κέντρον ἐπὶ τῆς ΔΒ, καὶ ἔσται δηλαδὴ τὸ ΑΒΓ τμῆμα μεῖζον ἡμικυκλίου.
But if the angle ΑΒΔ be less than ΒΑΔ, and we construct an angle equal to the angle ΑΒΔ on the straight line ΒΑ and at the point Α on it, the center will fall on ΔΒ within the segment ΑΒΓ, and obviously the segment ΑΒΓ will be greater than a semicircle.
κύκλου ἄρα τμήματος δοθέντος προσαναγέγραπται ὁ κύκλος· ὅπερ ἔδει ποιῆσαι.
Therefore, given a segment of a circle, the circle has been described; which was to be done.