a nugatory principle.' Now let us see, mein Herr, whether Mr. Cooley has done anything worthy of the writer of such brave 'orts' (as Shakespeare has it): and first let me ask how you define Parallel Lines.

NIEMAND reads.

'Right Lines are said to be parallel when they are equally and similarly inclined to the same right Line, or make equal angles with it towards the same side.'

Min. That is to say, if we see a Pair of Lines cut by a certain transversal, and are told that they make equal angles with it, we say 'these Lines are parallel'; and conversely, if we are told that a Pair of Lines are parallel, we say then there is a transversal, somewhere, which makes equal angles with them '?

Nie. Surely, surely.

Min. But we have no means of finding it? We have no right to draw a transversal at random and say 'this is the one which makes equal angles with the Pair'?

Nie. Ahem! Ahem! Ahem!

Min. You seem to have a bad cough.

Nie. Let us go to the next subject.

Min. Not till you have answered my question. Have we any means of finding the particular transversal which makes the equal angles?

Nie. I am sorry for my client, but, since you are so exigeant, I fear I must confess that we have no means of finding it.

Min. Now for your proof of Euc. I. 32.

Nie. You will allow us a preliminary Theorem?

Min. As many as you like.

Nie. Well, here is our Theorem II.

When two parallel

straight Lines AB, CD, are cut by a third straight Line EF, they make with it the alternate angles AGH, GHD, equal; and also the two internal angles at the same side BGH, GHD equal to two right angles.

A

E

D

For AGH and EGB are equal because vertically opposite, and EGB is also equal to GHD (Definition); therefore—' Min. There I must interrupt you. How do you know that EGB is equal to GHD? I grant you that, by the Definition, AB and CD make equal angles with a certain transversal: but have you any ground for saying that EF is the transversal in question?

Nie. We have not. We surrender at discretion. You will permit us to march out with the honours of war?

Min. We grant it you of our royal grace. March him off the table, and bring on the next Rival.

ACT II.

SCENE IV.

Treatment of Parallels by equidistances.

CUTHBERTSON.

Thou art so near, and yet so far.'

Modern Song.

Nie. I now lay before you Euclidian Geometry,' by FRANCIS CUTHBERTSON, M.A., late Fellow of C. C. C., Cambridge; Head Mathematical Master of the City of London School; published in 1874.

Min. It will not be necessary to discuss with you all the innovations of Mr. Cuthbertson's book. The questions of the separation of Problems and Theorems, the use of superposition, and the omission of the diagonals in Book II, are general questions which I have considered by themselves. The only points, which you and I need consider, are the methods adopted in treating Right Lines, Angles, and Parallels, wherever those methods differ from Euclid's.

The first subject, then, is the Right Line. How do you define and test it?

Nie. As in Euclid. But we prove what Euclid has assumed as an Axiom, namely, that two right Lines cannot have a common segment.

Min. I am glad to hear you assert that Euclid has assumed it as an Axiom,' for the interpolated and illogical corollary to Euc. I. 11 has caused many to overlook the fact that he has assumed it as early as Prop. 4, if not in Prop. 1. What is your proof?

NIEMAND reads.

'Two straight Lines cannot have a common segment.'

B

H

For if two straight Lines ABC, ABH could have a common segment AB; then the straight Line ABC might be turned about its extremity A, towards the side on which BH is, so as to cut BH; and thus two straight Lines would enclose a space, which is impossible.'

Min. You assume that, before C crosses BH, the portions coinciding along AB will diverge. But, if ABH is a right Line, this will not happen till C has passed H.

Nie. But you would then have one portion of the revolving Line in motion, and another portion at rest. Min. Well, why not?

Nie. We may assume that to be impossible; and that,

F

if a Line revolves about its extremity, it all moves at

once.

Min. Which, I take the liberty to think, is quite as great an assumption as Euclid's. I think the Axiom quite plain enough without any proof.

Your treatment of angles, and right angles, is the same as Euclid's, I think?

Nie. Yes, except that we prove that all right angles are equal.'

Min. Well, it is capable of proof, and therefore had better not be retained as an Axiom.

I must now ask you to give me your proof of Euc. I. 32.

Nie. We prove as far as I. 28 as in Euclid. In order to prove I. 29, we first prove, as a Corollary to Euc. I. 20, that 'the shortest distance between two points is a straight Line.'

Min. What is your next step?

Nie. A Problem (Pr. F. p. 52) in which we prove the Theorem that, of all right Lines drawn from a point to a Line, the perpendicular is the least.

Min. We will take that as proved.

Nie. We then deduce that the perpendicular is the shortest path from a point to a Line.

Next comes a Definition.

By the distance of a point from a straight Line is meant the shortest path from the point to the Line.'

Min. Have you anywhere defined the distance of one point from another?

Nie. No.