EUCLID IN REAL LIFE.
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.
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.
In the same way you can go on planting the old tree all over the bed—anywhere you like. In every case you'll get those right angles in the same ridiculous position—why, it makes me laugh now to think of it—and you'll be brought back to dear old CE.
And, of course, any point in CE except F would divide CE unequally, which I notice now is just what you've done yourself; so F is wrong too.
But you see the idea?
What a mess you've made of the bed!
Book I., Proposition 20.
Theorem.—Any two sides of a triangle are together greater than the third side.
Let ABC be a triangle.
Construction.—You know the eleventh hole? Well, let B be the tee, and let C be the green, and let BC be my drive. Yes, mine. Is it dead? Yes.
Now let BA be your drive. I'm afraid you've pulled it a bit and gone into the road by the farm.
You can't get on to the green by the direct route AC because you're under the wall. You'll have to play further up the road till you get opposite that gap at D. It's a pity, because you'll have to play about the same distance, only in the wrong direction.
Take your niblick, then, and play your second, making AD equal to AC. Now join CD.
I mean, put your third on the green. You can do that, surely? Good.
Proof.—There, I'm down in two. But we won't rub it in. Do you notice anything odd about these triangles? No? Well, the fact is that AD is equal to AC, and the result of that is that the angle ACD is equal to the angle ADC. That's Prop. 5. Anyhow, it's obvious, isn't it?
But steady on. The angle BCD is greater than its part, the angle ACD—you must admit that? (Look out, there's a fellow going to drive.)
And therefore the angle BCD—Oh, well, I can't go into it all now or it will mean we shall have to let these people through; but if you carry on on those lines you'll find that BD is greater than BC.
I mean you've only got to go back to where you played your third and you'll see that it must be so, won't you? Very well, then, don't argue.
But BD is equal to BA and AC, for AD is equal to AC; it had to be, you remember.
Therefore—now follow this closely—the two sides BA and AC are together greater than the third side BC.
That means, you see, that by pulling your drive out to the left there you gave yourself a lot of extra distance to cover.
You'd never have guessed that, would you? But old Euclid did.
Come along, then; they're putting. You must be more careful at this hole.
I think it's that right shoulder of yours...
A. P. H.