Let $a,b,c,d$ be the side lengths and $e,f$ the diagonal lengths of quadrilateral $ABCD$ with $E$ the intersection of the diagonals, and $\phi$ the acute angle between the diagonals. We have $AB^2=AE^2+BE^2-2AE*BE*cos\phi$, etc. Adding and subtracting yields that $a^2+c^2-b^2-d^2= 2ef \cos\phi$. But $ac+bd\ge ef$, with equality iff the quadrilateral is cyclic. So $\cos \phi \ge \frac{a^2+c^2-b^2-d^2}{2(ac+bd)}$, and so given $a,b,c,d$, $\phi$ is maximized by the unique quadrilateral which is cyclic.