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.