Does this proof work to prove that the greatest area of a triangle inside a circle is when the triangle is equilateral? Does this proof work to prove that the greatest area of a triangle inside a circle is when the triangle is equilateral? I gather it doesn't because most of the proofs I've seen use derivatives etc. If so why doesn't it work?
Consider a triangle $ABC$ inscribed inside a circle $\Gamma$. Assume WLOG one of the sided is fixed then one can easily see that the other two sides being equal maximizes the area of the triangle (it has the greatest height). Now consider one of the two equal sides - using that side as a base we can see from the same argument as before that the triangle must in fact be equilateral to maximize the area (this triangle has the greatest height as before).
Thanks in advance.
 A: You have proven that as long as the three sides are not all equal the triangle does not have maximal area. If you assume without proof that a maximal triangle exists you are done. But note that the following could still be true: There is no maximal triangle at all, and someone comes up with some method enlarging even an equilateral triangle. 
In order to prove that there is in fact a maximal triangle one can use a compactness argument: The area of the triangle with vertices $re^{i\phi_k}$ $\>(1\leq k\leq3)$ is a continuous function on $[0,2\pi]^3$; therefore it assumes a maximum for certain choices of the $\phi_k$.
A completely elementary proof that the equilateral triangle has maximal area is not so easy. You can find one (in German) here, pp. 60/61:
http://www.math.ethz.ch/~blatter/SolutionsI.pdf
A: Let us some analytic geometry using your idea: suppose our circle is $\,x^2+y^2=R^2\;$ , and we're going to fix, as you did, one of our sides as being parallel to the $\,x-$axis for simplicity, so that its end points are $\,A=(a,b)\;,\;\;B=(-a,b)\;,\;\;a>0\;,\;a^2+b^2=R^2\;,\;\;b\le 0\;$ . 
Let the other vertex be  $\,C=(0,R)\;$ since, as you remarked, for the above two points of the triangle, the maximal height is obtained when  the other two sides are equal, which means the third vertex is on the sides perpendicular bisector, which is easy to see is the $\,y-$axis. 
Thus, the area of the triangle depends on the distance $\,R-b=R-\sqrt{R^2-a^2}\,$ and on the horizonal side length's $\,2a\;$ :
$$f(a):=a\left(R-\sqrt{R^2-a^2}\right)\implies f'(a)=R-\sqrt{R^2-a^2}+\frac{a^2}{\sqrt{R^2-a^2}}\stackrel ?=0\iff$$
$$R\sqrt{R^2-a^2}-R^2+a^2+a^2=0\implies(2a^2-R^2)^2=R^2(R^2-a^2)\implies$$
$$R^4-4R^2a^2+4a^4=R^4-R^2a^2\implies a^2\left(4a^2-3R^2\right)=0\implies$$
$$\implies a=\frac{\sqrt3}2R$$ 
so that the slope of the line 
$$CA\,\;,\;\;A=\left(\frac{\sqrt3}2R\,,\,-\frac12R\right)\;,\;C=(0,R)\;\,\;\;\text{is}$$
$$-\frac{\frac32R}{\frac{\sqrt3}2R}=-\sqrt3=\tan\frac{-\pi}3$$
and we're done...
