Euclid · Elements §3.prop.9-3.prop.10

Criterion for the Center and Intersection Points of Circles

Passage 41 of 316 · Greek

Summary

In Proposition 9, it is proved that if three or more equal straight lines fall from an interior point to the circumference, that point is the center. In Proposition 10, it is proved by contradiction that two circles cannot intersect at more than two points.

§3.prop.9ἐὰν κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους ἢ δύο ἴσαι εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου.

If a point be taken inside a circle, and from the point there fall on the circle more than two equal straight lines, the point taken is the center of the circle.

ἔστω κύκλος ὁ ΑΒΓ, ἐντὸς δὲ αὐτοῦ σημεῖον τὸ Δ, καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν πλείους ἢ δύο ἴσαι εὐθεῖαι αἱ ΔΑ, ΔΒ, ΔΓ· λέγω, ὅτι τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου.

Let ΑΒΓ be a circle, and Δ a point inside it, and from Δ let there fall on the circle ΑΒΓ more than two equal straight lines, namely ΔΑ, ΔΒ, ΔΓ; I say that the point Δ is the center of the circle ΑΒΓ.

ἐπεζεύχθωσαν γὰρ αἱ ΑΒ, ΒΓ καὶ τετμήσθωσαν δίχα κατὰ τὰ Ε, Ζ σημεῖα, καὶ ἐπιζευχθεῖσαι αἱ ΕΔ, ΖΔ διήχθωσαν ἐπὶ τὰ Η, Κ, Θ, Λ σημεῖα.

For let ΑΒ, ΒΓ be joined and bisected at the points Ε, Ζ, and let ΕΔ, ΖΔ, being joined, be drawn through to the points Η, Κ, Θ, Λ.

ἐπεὶ οὖν ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, κοινὴ δὲ ἡ ΕΔ, δύο δὴ αἱ ΑΕ, ΕΔ δύο ταῖς ΒΕ, ΕΔ ἴσαι εἰσίν· καὶ βάσις ἡ ΔΑ βάσει τῇ ΔΒ ἴση· γωνία ἄρα ἡ ὑπὸ ΑΕΔ γωνίᾳ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν·

Since therefore ΑΕ is equal to ΕΒ, and ΕΔ is common, therefore the two ΑΕ, ΕΔ are equal to the two ΒΕ, ΕΔ; and the base ΔΑ is equal to the base ΔΒ; therefore the angle ΑΕΔ is equal to the angle ΒΕΔ.

ὀρθὴ ἄρα ἑκατέρα τῶν ὑπὸ ΑΕΔ, ΒΕΔ γωνιῶν· ἡ ΗΚ ἄρα τὴν ΑΒ τέμνει δίχα καὶ πρὸς ὀρθάς.

Therefore each of the angles ΑΕΔ, ΒΕΔ is right; therefore ΗΚ cuts ΑΒ in half and at right angles.

καὶ ἐπεί, ἐὰν ἐν κύκλῳ εὐθεῖά τις εὐθεῖάν τινα δίχα τε καὶ πρὸς ὀρθὰς τέμνῃ, ἐπὶ τῆς τεμνούσης ἐστὶ τὸ κέντρον τοῦ κύκλου, ἐπὶ τῆς ΗΚ ἄρα ἐστὶ τὸ κέντρον τοῦ κύκλου.

And since, if in a circle a straight line cut another straight line in half and at right angles, the center of the circle is on the cutting line, therefore the center of the circle is on ΗΚ.

διὰ τὰ αὐτὰ δὴ καὶ ἐπὶ τῆς ΘΛ ἐστι τὸ κέντρον τοῦ ΑΒΓ κύκλου.

For the same reasons, then, the center of the circle ΑΒΓ is also on ΘΛ.

καὶ οὐδὲν ἕτερον κοινὸν ἔχουσιν αἱ ΗΚ, ΘΛ εὐθεῖαι ἢ τὸ Δ σημεῖον· τὸ Δ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου.

And the straight lines ΗΚ, ΘΛ have no other common point than the point Δ; therefore the point Δ is the center of the circle ΑΒΓ.

ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους ἢ δύο ἴσαι εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου· ὅπερ ἔδει δεῖξαι.

If therefore a point be taken inside a circle, and from the point there fall on the circle more than two equal straight lines, the point taken is the center of the circle; which was to be proved.

§3.prop.10κύκλος κύκλον οὐ τέμνει κατὰ πλείονα σημεῖα ἢ δύο.

A circle does not cut a circle at more points than two.

εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓ κύκλον τὸν ΔΕΖ τεμνέτω κατὰ πλείονα σημεῖα ἢ δύο τὰ Β, Η, Ζ, Θ, καὶ ἐπιζευχθεῖσαι αἱ ΒΘ, ΒΗ δίχα τεμνέσθωσαν κατὰ τὰ κ, Λ σημεῖα· καὶ ἀπὸ τῶν Κ, Λ ταῖς ΒΘ, ΒΗ πρὸς ὀρθὰς ἀχθεῖσαι αἱ ΚΓ, ΛΜ διήχθωσαν ἐπὶ τὰ Α, Ε σημεῖα.

For, if possible, let a circle ΑΒΓ cut a circle ΔΕΖ at more points than two, namely Β, Η, Ζ, Θ, and let ΒΘ, ΒΗ, being joined, be bisected at the points Κ, Λ; and let ΚΓ, ΛΜ, being drawn from Κ, Λ at right angles to ΒΘ, ΒΗ, be carried through to the points Α, Ε.

ἐπεὶ οὖν ἐν κύκλῳ τῷ ΑΒΓ εὐθεῖά τις ἡ ΑΓ εὐθεῖάν τινα τὴν ΒΘ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΑΓ ἄρα ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου.

Since therefore in the circle ΑΒΓ a straight line ΑΓ cuts a straight line ΒΘ in half and at right angles, therefore the center of the circle ΑΒΓ is on ΑΓ.

πάλιν, ἐπεὶ ἐν κύκλῳ τῷ αὐτῷ τῷ ΑΒΓ εὐθεῖά τις ἡ ΝΞ εὐθεῖάν τινα τὴν ΒΗ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΝΞ ἄρα ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου.

Again, since in the same circle ΑΒΓ a straight line ΝΞ cuts a straight line ΒΗ in half and at right angles, therefore the center of the circle ΑΒΓ is on ΝΞ.

ἐδείχθη δὲ καὶ ἐπὶ τῆς ΑΓ, καὶ κατʼ οὐδὲν συμβάλλουσιν αἱ ΑΓ, ΝΞ εὐθεῖαι ἢ κατὰ τὸ Ο· τὸ Ο ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου.

But it was also proved to be on ΑΓ, and the straight lines ΑΓ, ΝΞ meet at no other point than at Ο; therefore the point Ο is the center of the circle ΑΒΓ.

ὁμοίως δὴ δείξομεν, ὅτι καὶ τοῦ ΔΕΖ κύκλου κέντρον ἐστὶ τὸ Ο· δύο ἄρα κύκλων τεμνόντων ἀλλήλους τῶν ΑΒΓ, ΔΕΖ τὸ αὐτό ἐστι κέντρον τὸ Ο· ὅπερ ἐστὶν ἀδύνατον.

Similarly then we shall prove that Ο is also the center of the circle ΔΕΖ; therefore two intersecting circles ΑΒΓ, ΔΕΖ have the same center Ο; which is impossible.

οὐκ ἄρα κύκλος κύκλον τέμνει κατὰ πλείονα σημεῖα ἢ δύο· ὅπερ ἔδει δεῖξαι.

Therefore a circle does not cut a circle at more points than two; which was to be proved.

Notes

§3.prop.9πλείους ἢ δύο — The phrase "more than two" (πλείους ἢ δύο) in this geometric context practically means "three or more." While two equal straight lines can be drawn to the circumference from any point other than the center, the existence of three or more equal lines uniquely identifies the point as the center.
¦15¦ἡ ΝΞ — In the preceding construction, the perpendicular drawn from the point Λ was named "ΛΜ" and extended to "Α, Ε," but here the straight line is suddenly referred to as "ΝΞ." This suggests either a discrepancy in the manuscript diagrams or a reference to the full line extending "ΛΜ." Syntactically, it functions as the perpendicular bisector of "ΒΗ."

Cite this passage

Euclid, Elements §3.prop.9-3.prop.10. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:3.prop.9-3.prop.10

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