Thales theorem to show half-circle encloses largest area over a line given a circumference I'm working on a proof for the isoperimetric inequality, and want to show that for all curves over a line, the half-circle encloses the maximum area with that line given a fixed circumference.
I was given a hint to use Thales's theorem.
The hint confuses me more than it helps me, so I was hoping that someone might be able to shed some light on it. What I don't get is why the right triangles from Thales's theorem are useful if they are obviously not disjoint, and I don't think I can use them to approximate the area.
Does someone see the relationship?
 A: You should first review this page:
http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml
Specifically, the section beginning with "Proof of Statement 1" contains Lemma 2, which is the part you are working on.
For the sake of completeness, I have reproduced it below.

Lemma 2.  Consider all the arcs with a given length and the endpoints $S$ and $T$ on a fixed straight line. The curve that encloses the maximum area between the curve and the straight line is a semicircle.


Proof.  Suffice it to show that every angle inscribed into the arc is right. If there is a point on the arc, say $P$, for which $\angle SPT$ is not right, slide either point $S$ or point $T$ along the straight line $ST$ until the angle becomes right. Let pieces of the arc move along. As an exercise prove that among all triangles with two given sides the one whose sides enclose a right angle has the largest area. Since the area of the two red regions did not change but the area of the triangle grew, the whole area between the new curve $SPT$ and the line $ST$ has increased. This shows that unless the curve is a semicircle we can always increase the area in question by moving points $S$ and $T$. This proves Lemma 2 and with it Statement 1.

However, note the remark that follows this proof, in which it is pointed out that there is a flaw in the line of reasoning for Statement 1.  This does not invalidate the proof of Lemma 2 which you are specifically interested in, but it does mean that the argument in which Lemma 2 plays a role is incomplete.
The missing piece in Lemma 2 is the "exercise" which is actually quite simple to show, which is why I have omitted the proof.
