If it was not for the paper-shortage I should at once re-write EUCLID, or those parts of him which I understand. The trouble about old EUCLID was that he had no soul, and few of his books have that emotional appeal for which we look in these days. My aim would be to bring home his discoveries to the young by clothing them with human interest; and I should at the same time demonstrate to the adult how often they might be made practically useful in everyday life. When one thinks of the times one draws a straight line at right angles to another straight line, and how seldom one does it EUCLID'S way ... every time one writes a T.... Well, let us take, for example-- BOOK III., PROPOSITION 1. PROBLEM.--_To find the centre of a given circle_. Let ABC be that horrible round bed where you had the geraniums last year. This year, I gather, the idea is to have it nothing but rose-trees, with a great big fellow in the middle. The question is, where is the middle? I mean, if you plant it in a hurry on your own judgment, everyone who comes near the house will point out that the bed is all cock-eye. Besides, you can see it from the dining-room and it will annoy you at breakfast. [Illustration] CONSTRUCTION.--Well, this is how we go about it. First, you draw any chord AB in the given bed ABC. You can do that with one of those long strings the gardener keeps in his shed, with pegs at the end. Bisect AB at D. Now don't look so stupid. We've done that already in Book I., Prop. 10, you remember, when we bisected the stick of nougat. That's right. Now from D draw DC at right angles to AB, and meeting the lawn at C. You can do that with a hoe. Produce CD to meet the lawn again at E. Now we do some more of that bisecting; this time we bisect EC at F. Then F shall be the middle of the bed; and that's where your rose-tree is going. PROOF???--Well, I mean, if F be _not_ the centre let some point G, outside the line CE, be the centre and put the confounded tree _there_. And, what's more, you can jolly well join GA, GD and GB, and see what that looks like. Just cast your eye over the two triangles GDA and GDB. Don't you see that DA is equal to DB (unless, of course, you've bisected that chord all wrong), and DG is common, and GA is equal to GB--at least according to your absurd theory about G it is, since they must be both _radii_. _Radii_ indeed! _Look_ at them. Ha, ha! Therefore, you fool, the angle GDA is equal to the angle GDB. Therefore they are both right angles. Therefore the angle GDA is a right angle. (I know you think I'm repeating myself, but you'll see what I'm getting at in a minute.) _Therefore_--and this is the cream of the joke--therefore--really, I can't help laughing--therefore _the angle CDA is equal to the angle GDA!_ That is, the part is equal to the whole--which is ridiculous. I mean, it's too _laughable_. So, you see, your rose-tree is not in the middle at all.