Clay Mathematics Institute

Dedicated to increasing and disseminating mathematical knowledge

Book 3 Proposition 1



Τοῦ δοθέντος κύκλου τὸ κέντρον εὑρεῖν. Ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ: δεῖ δὴ τοῦ ΑΒΓ κύκλου τὸ κέντρον εὑρεῖν. 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: λέγω, ὅτι τὸ Ζ κέντρον ἐστὶ τοῦ ΑΒΓ [κύκλου]. Μὴ γάρ, ἀλλ' 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. ὅταν δὲ εὐθεῖα ἐπ' εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν: 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. οὐκ ἄρα τὸ Η κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδ' ἄλλο τι πλὴν τοῦ Ζ. Τὸ Ζ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ [κύκλου].

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.

To find the centre of a given circle. Let ABC be the given circle; thus it is required to find the centre of the circle ABC. Let a straight line AB be drawn through it at random, and let it be bisected at the point D; from D let DC be drawn at right angles to AB and let it be drawn through to E; let CE be bisected at F; I say that F is the centre of the circle ABC. For suppose it is not, but, if possible, let G be the centre, and let GA, GD, GB be joined. Then, since AD is equal to DB, and DG is common, the two sides AD, DG are equal to the two sides BD, DG respectively; and the base GA is equal to the base GB, for they are radii; therefore the angle ADG is equal to the angle GDB. [I. 8] But, 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; [I. Def. 10] therefore the angle GDB is right. But the angle FDB is also right; therefore the angle FDB is equal to the angle GDB, the greater to the less: which is impossible. Therefore G is not the centre of the circle ABC. Similarly we can prove that neither is any other point except F. Therefore the point F is the centre of the circle ABC.

Porism. From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line.