§3.prop.3ἐὰν ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει· καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει.
If in a circle a straight line through the center bisect a straight line not through the center, it also cuts it at right angles; and if it cut it at right angles, it also bisects it.
ἔστω κύκλος ὁ ΑΒΓ, καὶ ἐν αὐτῷ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΓΔ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΒ δίχα τεμνέτω κατὰ τὸ Ζ σημεῖον· λέγω, ὅτι καὶ πρὸς ὀρθὰς αὐτὴν τέμνει.
Let ABC be a circle, and in it let a straight line GD through the center bisect a straight line AB not through the center at the point Z; I say that it also cuts it at right angles.
εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Ε, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΕ, δύο δυσὶν ἴσαι.
For let the center of the circle ABC be taken, and let it be E, and let EA, EB be joined. And since AZ is equal to ZB, and ZE is common, the two are equal to the two.
καὶ βάσις ἡ ΕΑ βάσει τῇ ΕΒ ἴση· γωνία ἄρα ἡ ὑπὸ ΑΖΕ γωνίᾳ τῇ ὑπὸ ΒΖΕ ἴση ἐστίν.
And the base EA is equal to the base EB; therefore the angle AZE is equal to the angle BZE.
ὅταν δὲ εὐθεῖα ἐπʼ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν· ἑκατέρα ἄρα τῶν ὑπὸ ΑΖΕ, ΒΖΕ ὀρθή 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 each of the angles AZE, BZE is right.
ἡ ΓΔ ἄρα διὰ τοῦ κέντρου οὖσα τὴν ΑΒ μὴ διὰ τοῦ κέντρου οὖσαν δίχα τέμνουσα καὶ πρὸς ὀρθὰς τέμνει.
Therefore GD, which is through the center, cutting AB, which is not through the center, bisecting it, also cuts it at right angles.
ἀλλὰ δὴ ἡ ΓΔ τὴν ΑΒ πρὸς ὀρθὰς τεμνέτω· λέγω, ὅτι καὶ δίχα αὐτὴν τέμνει, τουτέστιν, ὅτι ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ.
τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴση ἐστὶν ἡ ΕΑ τῇ ΕΒ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΕΑΖ τῇ ὑπὸ ΕΒΖ. ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΖΕ ὀρθῇ τῇ ὑπὸ ΒΖΕ ἴση· δύο ἄρα τρίγωνά ἐστι τὰ ΕΑΖ, ΕΖΒ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΕΖ ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει· ἴση ἄρα ἡ ΑΖ τῇ ΖΒ.
ἐὰν ἄρα ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει·
But now let GD cut AB at right angles; I say that it also bisects it, that is, that AZ is equal to ZB. For, with the same construction, since EA is equal to EB, the angle EAZ is also equal to the angle EBZ. And the right angle AZE is also equal to the right angle BZE; therefore there are two triangles EAZ, EZB having two angles equal to two angles, and one side equal to one side, namely their common side EZ subtending one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; therefore AZ is equal to ZB.
καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει· ὅπερ ἔδει δεῖξαι.
If then in a circle a straight line through the center bisect a straight line not through the center, it also cuts it at right angles; and if it cut it at right angles, it also bisects it; which was meet to show.
§3.prop.4ἐὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ διὰ τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα.
If in a circle two straight lines cut one another, not being through the center, they do not bisect one another.
ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε μὴ διὰ τοῦ κέντρου οὖσαι· λέγω, ὅτι οὐ τέμνουσιν ἀλλήλας δίχα.
Let ABCD be a circle, and in it let two straight lines AC, BD cut one another at the point E, not being through the center; I say that they do not bisect one another.
εἰ γὰρ δυνατόν, τεμνέτωσαν ἀλλήλας δίχα ὥστε ἴσην εἶναι τὴν μὲν ΑΕ τῇ ΕΓ, τὴν δὲ ΒΕ τῇ ΕΔ· καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΖΕ.
ἐπεὶ οὖν εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΖΕ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ δίχα τέμνει, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει· ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΕΑ·
For, if possible, let them bisect one another, so that AE is equal to EC, and BE to ED; and let the center of the circle ABCD be taken, and let it be Z, and let ZE be joined. Since then a straight line ZE through the center bisects a straight line AC not through the center, it also cuts it at right angles; therefore the angle ZEA is right.
πάλιν, ἐπεὶ εὐθεῖά τις ἡ ΖΕ εὐθεῖάν τινα τὴν ΒΔ δίχα τέμνει, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει· ὀρθὴ ἄρα ἡ ὑπὸ ΖΕΒ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΖΕΑ ὀρθή· ἴση ἄρα ἡ ὑπὸ ΖΕΑ τῇ ὑπὸ ΖΕΒ ἡ ἐλάττων τῇ μείζονι· ὅπερ ἐστὶν ἀδύνατον.
Again, since a straight line ZE bisects a straight line BD, it also cuts it at right angles; therefore the angle ZEB is right. But the angle ZEA was also proved right; therefore the angle ZEA is equal to the angle ZEB, the less to the greater: which is impossible.
οὐκ ἄρα αἱ ΑΓ, ΒΔ τέμνουσιν ἀλλήλας δίχα.
Therefore AC, BD do not bisect one another.
ἐὰν ἄρα ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ διὰ τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα· ὅπερ ἔδει δεῖξαι.
If then in a circle two straight lines cut one another, not being through the center, they do not bisect one another; which was meet to show.
§3.prop.5ἐὰν δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον.
If two circles cut one another, they will not have the same center.
δύο γὰρ κύκλοι οἱ ΑΒΓ, ΓΔΗ τεμνέτωσαν ἀλλήλους κατὰ τὰ Β, Γ σημεῖα. λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον.
For let two circles ABC, CDH cut one another at the points B, G; I say that they will not have the same center.
εἰ γὰρ δυνατόν, ἔστω τὸ Ε, καὶ ἐπεζεύχθω ἡ ΕΓ, καὶ διήχθω ἡ ΕΖΗ, ὡς ἔτυχεν.
For, if possible, let it be E, and let EG be joined, and let EZH be drawn at random.
καὶ ἐπεὶ τὸ Ε σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΕΓ τῇ ΕΖ. πάλιν, ἐπεὶ τὸ Ε σημεῖον κέντρον ἐστὶ τοῦ ΓΔΗ κύκλου, ἴση ἐστὶν ἡ ΕΓ τῇ ΕΗ·
And since the point E is the center of the circle ABC, EG is equal to EZ. Again, since the point E is the center of the circle CDH, EG is equal to EH.
ἐδείχθη δὲ ἡ ΕΓ καὶ τῇ ΕΖ ἴση· καὶ ἡ ΕΖ ἄρα τῇ ΕΗ ἐστιν ἴση ἡ ἐλάσσων τῇ μείζονι· ὅπερ ἐστὶν ἀδύνατον.
But EG was also proved equal to EZ; therefore EZ is also equal to EH, the less to the greater: which is impossible.
οὐκ ἄρα τὸ Ε σημεῖον κέντρον ἐστὶ τῶν ΑΒΓ, ΓΔΗ κύκλων.
Therefore the point E is not the center of the circles ABC, CDH.
ἐὰν ἄρα δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔστιν αὐτῶν τὸ αὐτὸ κέντρον· ὅπερ ἔδει δεῖξαι.
If then two circles cut one another, they will not have the same center; which was meet to show.