Commentary on Proposition III.10

Back to Proposition III.10.

There is nothing in the demonstration of this proposition which assumes that the circles cut one another; it proves that two circles cannot meet at more than two points, whether they cut or meet without cutting, i.e. touch one another.

Here again, of two demonstrations given in the earlier texts, Simson chose the second, which August and Heiberg relegate to an Appendix and which is as follows: “For again let the circle ABC cut the circle DEF at more points than two, namely B, G, H, F; let the centre K of the circle ABC be taken, and let KB, KG, KF be joined.

Since then a point K has been taken within the circle DEF, and from K more than two straight lines, namely KB, KF, KG, have fallen on the circle DEF,

[III. 9]the point K is the centre of the circle DEF.
A B C D E F G H K

But $K$ is also the centre of the circle ABC.

Therefore two circles cutting one another have the same centre K: which is impossible. [III. 5]

Therefore a circle does not cut a circle at more points than two.
Q. E. D."

This demonstration is claimed by Heron (see an-Nairīzī, ed. Curtze, pp. 120—1). It is incomplete because it assumes that the point K which is taken as the centre of the circle ABC is within the circle DEF. It can however be completed by means of III. 8 and the corresponding proposition with reference to a point on the circumference of a circle which was enunciated in the note on III. 8. For (1) if the point K is on the circumference of the circle DEF, we obtain a contradiction of the latter proposition which asserts that only two equal straight lines can be drawn from K to the circumference of the circle DEF; (2) if the point K is outside the circle DEF, we obtain a contradiction of the corresponding part of III. 8.

Euclid’s proof contains an unproved assumption, namely that the lines bisecting BG, BH at right angles will meet in a point P. For a discussion of this assumption see note on IV. 5.