Euc. Thanks: it is rather a strain on the imaginative faculty.
Min. You mean, then, that the multiple of an angle may be conceived of as so many separate angles, not in contact, nor added together into one?
Euc. Certainly.
Min. But you have to contemplate the case where two such angular magnitudes are equal, and to infer from that, by III. 26, that the subtending arcs are equal. How can you infer this when your angular magnitude is not one angle but many?
Euc. Why, the sum total of the first set of angles is equal to the sum total of the second set. Hence the second set can clearly be broken up and put together again in such amounts as to make a set equal, each to each, to the first set: and then the sum total of the arcs, and likewise of the sectors, will evidently be equal also.
But if you contemplate the multiples of the angles as single angular magnitudes, I do not see how you prove the equality of the subtending arcs: for my proof applies only to cases where the angle is less than the sum of two right angles.
Min. That is very true, and you have quite convinced me that we ought to observe that limit, and not contemplate 'angles of rotation' till we enter on the subject of Trigonometry.
As to right angles, it has been suggested that your Axiom 'all right angles are equal to one another' is capable of proof as a Theorem.
Euc. I do not object to the interpolation of such a
