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.

[Illustration]

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.