At p. 19 of the 2nd Edition, the new Axiom, on which my Theory rests, (which re-appears, in a modified form, at p. 14 of the 3rd Edition), stood thus:—"In every Circle, the inscribed equilateral Hexagon is greater than any one of the Segments which lie outside it." My controversy with the Athenæum, on this Axiom, shall also be given in the form of a dialogue. Athenæum. (Oct. 27, 1888.) "... a stronger objection, in our opinion, is the implied assumption of the possibility of the inscribed equilateral Hexagon, a possibility which is not demonstrated till we reach the fifteenth Proposition of Euclid's fourth book." Author. (In the Preface to the 2nd Edition, at p. xi). "But does it need demonstrating? May we not assume (1) that the Magnitude 'four right angles' contains 6-6ths of itself; (2) that it is theoretically possible to draw radii dividing it into these 6-6ths? Once grant me this, and I ask no more. I have then the logical right to join the ends of these radii, and to prove (by Euc. I. 4) that the chords are equal." [Pg xiii] Athenæum. (Oct. 5, 1889.) "We objected that it was not consistent with the spirit or practice of Euclid's reasoning to assume the 'theoretical possibility' of a regular Hexagon inscribed in a Circle, without first proving that such a figure could be actually constructed from his three postulates. Euclid's restrictions may be arbitrary, unnecessary, cramping, vexatious, absurd—indeed, we think they deserve these and many other epithets—but there they are, and, if Mr. Dodgson accepts them, he is bound to keep his assumptions within the boundaries which they prescribe." Author. "You're particular to a shade (as Scrooge said to Marley's ghost): however, I'll do what I can to oblige you. I presume you will be satisfied if I can, without using more of Euclid than his first 28 Propositions, construct an angle which shall be 1-6th of 4 right angles? Very good. First, then, with the help of his arbitrary Prop. I, I construct an equilateral Triangle. Next, by his unnecessary Prop. IX, I draw the bisectors of 2 of its angles. Next, by his cramping Post. 1, I join their point of intersection to the third vertex. Next, by his vexatious Prop. IV, I prove the 3 angles, whose common vertex is this point, to be equal. From which I draw the absurd conclusion that each of them is 1-3rd (and that therefore its half is 1-6th) of 4 right angles. How does that strike you?"