§3.prop.1τοῦ δοθέντος κύκλου τὸ κέντρον εὑρεῖν.
To find the center of a given circle.
ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ· δεῖ δὴ τοῦ ΑΒΓ κύκλου τὸ κέντρον εὑρεῖν.
Let the given circle be ABC; thus it is required to find the center of the circle ABC.
διήχθω τις εἰς αὐτόν, ὡς ἔτυχεν, εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω δίχα κατὰ τὸ Δ σημεῖον, καὶ ἀπὸ τοῦ Δ τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΔΓ καὶ διήχθω ἐπὶ τὸ Ε, καὶ τετμήσθω ἡ ΓΕ δίχα κατὰ τὸ Ζ·
Let any straight line AB be drawn through it at random, and let it be bisected at the point D, and from D let DG be drawn at right angles to AB and let it be produced to E, and let GE be bisected at Z; I say that Z is the center of the ABC.
λέγω, ὅτι τὸ Ζ κέντρον ἐστὶ τοῦ ΑΒΓ. μὴ γάρ, ἀλλʼ εἰ δυνατόν, ἔστω τὸ Η, καὶ ἐπεζεύχθωσαν αἱ ΗΑ, ΗΔ, ΗΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ ἡ ΔΗ, δύο δὴ αἱ ΑΔ, ΔΗ δύο ταῖς ΗΔ, ΔΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ βάσις ἡ ΗΑ βάσει τῇ ΗΒ ἐστιν ἴση· ἐκ κέντρου γάρ· γωνία ἄρα ἡ ὑπὸ ΑΔΗ γωνίᾳ τῇ ὑπὸ ΗΔΒ ἴση ἐστίν.
For, if not, let it be H, if possible, and let HA, HD, HB be joined. And since AD is equal to DB, and DH is common, the two AD, DH are equal to the two HD, DB respectively; and the base HA is equal to the base HB; for they are from the center; therefore the angle ADH is equal to the angle HDB.
ὅταν δὲ εὐθεῖα ἐπʼ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν·
And when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; therefore the angle HDB is right.
ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΗΔΒ. ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΔΒ ὀρθή· ἴση ἄρα ἡ ὑπὸ ΖΔΒ τῇ ὑπὸ ΗΔΒ, ἡ μείζων τῇ ἐλάττονι· ὅπερ ἐστὶν ἀδύνατον.
But the angle ZDB is also right; therefore the angle ZDB is equal to the angle HDB, the greater to the less: which is impossible.
οὐκ ἄρα τὸ Η κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου.
Therefore H is not the center of the circle ABC.
ὁμοίως δὴ δείξομεν, ὅτι οὐδʼ ἄλλο τι πλὴν τοῦ Ζ.
τὸ Ζ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ.
Similarly we can prove that neither is any other point except Z. Therefore the point Z is the center of the ABC.
Πόρισμα
ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἐν κύκλῳ εὐθεῖά τις εὐθεῖάν τινα δίχα καὶ πρὸς ὀρθὰς τέμνῃ, ἐπὶ τῆς τεμνούσης ἐστὶ τὸ κέντρον τοῦ κύκλου· ὅπερ ἔδει ποιῆσαι.
Porism From this it is manifest that, if in a circle a straight line cut a straight line bisecting it and at right angles, the center of the circle is on the cutting line; which was meet to do.
§3.prop.2ἐὰν κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου.
If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle.
ἔστω κύκλος ὁ ΑΒΓ, καὶ ἐπὶ τῆς περιφερείας αὐτοῦ εἰλήφθω δύο τυχόντα σημεῖα τὰ Α, Β· λέγω, ὅτι ἡ ἀπὸ τοῦ Α ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου.
Let ABC be a circle, and let two points A, B be taken at random on its circumference; I say that the straight line joined from A to B will fall within the circle.
μὴ γάρ, ἀλλʼ εἰ δυνατόν, πιπτέτω ἐκτὸς ὡς ἡ ΑΕΒ, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ δ, καὶ ἐπεζεύχθωσαν αἱ ΔΑ, ΔΒ, καὶ διήχθω ἡ ΔΖΕ.
ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΔΒ, ἴση ἄρα καὶ γωνία ἡ ὑπὸ ΔΑΕ τῇ ὑπὸ ΔΒΕ·
For, if not, let it fall without, as AEB, and let the center of the circle ABC be taken, and let it be D, and let DA, DB be joined, and let DZE be drawn. Since then DA is equal to DB, the angle DAE is also equal to the angle DBE.
καὶ ἐπεὶ τριγώνου τοῦ ΔΑΕ μία πλευρὰ προσεκβέβληται ἡ ΑΕΒ, μείζων ἄρα ἡ ὑπὸ ΔΕΒ γωνία τῆς ὑπὸ ΔΑΕ. ἴση δὲ ἡ ὑπὸ ΔΑΕ τῇ ὑπὸ ΔΒΕ·
And since one side AEB of the triangle DAE has been produced, therefore the angle DEB is greater than the angle DAE. But the angle DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE.
μείζων ἄρα ἡ ὑπὸ ΔΕΒ τῆς ὑπὸ ΔΒΕ. ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει·
And the greater side subtends the greater angle; therefore DB is greater than DE.
μείζων ἄρα ἡ ΔΒ τῆς ΔΕ. ἴση δὲ ἡ ΔΒ τῇ ΔΖ. μείζων ἄρα ἡ ΔΖ τῆς ΔΕ ἡ ἐλάττων τῆς μείζονος· ὅπερ ἐστὶν ἀδύνατον.
But DB is equal to DZ. Therefore DZ is greater than DE, the less than the greater: which is impossible.
οὐκ ἄρα ἡ ἀπὸ τοῦ Α ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἐκτὸς πεσεῖται τοῦ κύκλου.
Therefore the straight line joined from A to B will not fall without the circle.
ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἐπʼ αὐτῆς τῆς περιφερείας· ἐντὸς ἄρα.
Similarly we can prove that neither will it fall on the circumference itself; therefore within.
ἐὰν ἄρα κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου· ὅπερ ἔδει δεῖξαι.
If then on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle; which was meet to show.