I've finally had a chance to draw the diagram I couldn't quite visualize in enough detail while driving, and it works! I'm afraid my technical skills are not up to reproducing it online, so instead I'll give a wordy description and try to make it as clear as I can (but not too long).Lemma.
Draw a semicircle with the straight side horizontal and the semicircle above it. Let the two ends of the straight side be A and B, and let the top point be C. Let the midpoint of A and B be D. Let P be any point on the circle between C and B. Then the angle APC is 45 degrees.Proof.
It's a well known theorem in geometry that the angle is constant. When P=B, it is clearly 45 degrees.
Note also that the angle APB is 90 degrees, by the same well known theorem in geometry. So the angle you get by going from B to P and then directly away from C is also 45 degrees.
What is the relevance of this semicircle? Well, suppose we take another semicircle, also with a horizontal base, and take two points on it V and W such that the angle VOW (where O is the centre of the circle) is 90 degrees. Then the points V, O and W lie in a unique circle, and that circle cuts the base of the semicircle in two points, one of which is O. Let Q be the other point. I want to prove that the angle VQO is 45 degrees, as is the angle you get if you go from W to Q and then directly away from O. But VW is a diameter of the circle that goes through V, O and W (since the angle VOW is 90 degrees), so if we turn the diagram round and consider the semicircle that has VW as its base and goes through O and Q, then the lemma is telling us precisely what we want.
This is sort of doing things backwards: I'm showing that if the angle VOW is 90 degrees, then V and W will be the top points of two semicircles as in John's diagram. It is then not hard to prove uniqueness: that is, if 90 degrees works, then no other angle can work.
That may sound slightly complicated but if you draw the diagrams I describe, you'll see that it's an almost instant deduction from the lemma, which itself is an almost instant consequence of the well known geometrical theorem about the angle made by three points in a circle not depending on the middle point.
I forgot to say at the beginning of all this that I still need the Pythagoras step that converts the problem from one about areas to one about the length of the arc that joins the top points of the two semicircles. Maybe there's a nice proof that sees it done via areas rather than lengths. Now to look at +Stuart Presnell
's diagram to see whether it's the same idea.