Prove there is no maximum perimeter trapezoid inscribed in a circle I want to prove that there is no maximum perimeter trapezoid inscribed in a circle.  

Edit: the way my textbook defined a trapezoid does not allow for a square to fit in the trapezoid definition. If your definition is different than that, this problem is the equivalent of proving that the maximum perimeter trapezoid inscribed in a circle is a square.
What I have already figured:
I know (and know how to prove) that an inscribed trapezoid must be an isosceles trapezoid.
I also know that the maximum perimeter quadrilateral inscribed in a circle is a square (but don't know how to prove it).
I know that the inscribed trapezoid can get as close to being a square as we like, but it can never be a square. That is why there is no maximum perimeter inscribed trapezoid. But I do not know how to prove that the square is the maximum perimeter quadrilateral inscribed in a circle.
I think that I should approximate the inscribed square by an inscribed trapezoid, but when I tried that I couldn't prove that the perimeter of the square was greater than that of the trapezoid.
 A: I shall prove the statement in the title, given the book's definition of a trapezoid.
We may assume the trapezoid $T$ inscribed in the unit circle, with horizontal parallel sides $y=a$ and $y=b$, whereby $-1<a<b<1$, and $b>0$. If $a>0$ we can enlarge the perimeter of $T$ by replacing $a$ by $-a$. The four vertices of $T$ can then be written as
$$(\pm \cos\psi,-\sin\psi), \quad(\pm\cos\phi,\sin\phi)$$
with $0\leq\psi<{\pi\over2}$ and $0<\phi<{\pi\over2}$. The perimeter $L$ of $T$  computes to
$$L(\phi,\psi)=4\sin{\phi+\psi\over2}+2\cos\phi+2\cos\psi\ ,$$
such that
$$L_\phi=2\cos{\phi+\psi\over2}-2\sin\phi,\qquad L_\psi=2\cos{\phi+\psi\over2}-2\sin\psi\ .$$
If $\psi=0$ one has $L_\psi=2\cos{\phi\over2}>0$, hence increasing $\psi$ will increase $L$. If both $\phi$ and $\psi$ are $>0$ we can locally increase $L$ unless $L_\phi=L_\psi=0$. The latter would lead to $\psi=\phi$ and then to $\cos\phi=\sin\phi$, hence $\psi=\phi=45^\circ$, which is not feasible here. 
A: A trapezoid is a quadrilateral with at least one pair of opposite sides being parallel. A square fits this definition. Any theorem applying to a trapezoid also applies to a square.  
A: Hint: assume the convex cyclic quadrilateral $ABCD$ includes the center $O$ of its circumscribed circle or radius $r$ (otherwise find a suitable one with larger perimeter). Then each angle $\widehat{AOB}\le \pi$ and $\sum_{cyc}\widehat{AOB} = 2 \pi$. Given that the perimeter is $P = \sum_{cyc} 2 r \sin \widehat{AOB}/2$ and $\sin x$ is concave on $(0,\pi)$ it follows by Jensen's inequality that:
$$P/2r = \sum_{cyc} \sin \frac{\widehat{AOB}}{2} \le 4 \sin\left(\frac{1}{4} \sum_{cyc} \frac{\widehat{AOB}}{2}\right) = 4 \sin \frac{\pi}{4} = 2 \sqrt{2}$$
with equality iff all angles $\widehat{AOB}/2$ are equal i.e. the quadrilateral is a square.
A: Note the symmetries of the circle, you should be able to convince yourself that the maximum peremiter quadrilateral is one that maximizes symmetries (if there was a way to make the perimeter larger on one side of a line of symmetry the same action can always be taken by the other side) this leaves the square and the the crossed square as possible shapes, the crossed square has greater perimeter then the square but it is not the limit of expanding the symmetries of a trapezoid, therefore a square has the maximum perimeter of not self intersecting quadrilaterals.
