Three sides of a convex quadrilateral $ABCD$ have lengths $AB = a$, $BC = b$, and $CD = c$. If the area of the quadrilateral is as large as possible, prove that the length $x$ of the fourth side satisfies the equation $$x^3-(a^2+b^2+c^2)x-2abc=0$$
I know a solution by calculus, but I want to know if there is a solution by euclidean geometry and factorization (without derivatives).
Thanks for attention.
The calculus solution
$$d^2=a^2+b^2-2ab\cos(\theta)\\ Area=\frac{cd}{2}\sin\alpha+\frac{ab}{2}\sin(\theta)\\ d^2=x^2-c^2\\ x^2-c^2=a^2+b^2-2ab\cos(\theta)\\ \sin\theta=\frac{\sqrt{4a^2b^2-(x^2-a^2-b^2-c^2)^2}}{2ab}\\ Area=\frac{c.\sqrt{x^2-c^2}}{2}+\frac{ab}{2}.\frac{\sqrt{4a^2b^2-(x^2-a^2-b^2-c^2)^2}}{2ab}\\ Area=\frac{c.\sqrt{x^2-c^2}}{2}+\frac{\sqrt{4a^2b^2-(x^2-a^2-b^2-c^2)^2}}{4} $$
Doing $\frac{d}{dx}$
$$ \frac{2xc}{4\sqrt{x^2-c^2}}-\frac{2(x^2-a^2-b^2-c^2).2x}{8\sqrt{4a^2b^2-(x^2-a^2-b^2-c^2)^2}}=0\\ \frac{c}{\sqrt{x^2-c^2}}=\frac{(x^2-a^2-b^2-c^2)}{\sqrt{4a^2b^2-(x^2-a^2-b^2-c^2)^2}}\\ c\sqrt{4a^2b^2-(x^2-a^2-b^2-c^2)^2}=(x^2-a^2-b^2-c^2)\sqrt{x^2-c^2}\\ 4a^2b^2.c^2-c^2.(x^2-a^2-b^2-c^2)^2=(x^2-a^2-b^2-c^2)^2(x^2-c^2)\\ 4a^2b^2c^2=(x^2-a^2-b^2-c^2)^2.x^2\\ 2abc=(x^2-a^2-b^2-c^2).x\\ x^3-(a^2+b^2+c^2)x-2abc=0 $$
If the positions of sides $a$ and $b$ are fixed, maximum area is achieved when side $c$ is perpendicular diagonal $AC$. Analogously, if the positions of sides $b$ and $c$ are fixed, maximum area is achieved when side $a$ is perpendicular to diagonal $BD$. It follows that we have maximum area when $\angle ABD=\angle ACD=90°$, that is when quadrilateral $ABCD$ is inscribed in a half-circle of diameter $AD=x$.
To find $x$, let $2\alpha$, $2\beta$ and $2\gamma$ be the central angles subtending $a$, $b$ and $c$ respectively. We have then: $$ x={a\over\sin\alpha}={b\over\sin\beta}={c\over\sin\gamma}. $$ From the first equality we get $\sin\beta=(b/a)\sin\alpha$. From the second equality, taking into account that $\alpha+\beta+\gamma=90°$ and $\sin\gamma=\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$, we have: $$ a(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=c\sin\alpha, $$ that is: $$ a\sqrt{1-\sin^2\alpha}\sqrt{1-(b/a)^2\sin^2\alpha}=c\sin\alpha+b\sin^2\alpha. $$ Squaring both sides and rearranging we finally obtain: $$ a^2=(a^2+b^2+c^2)\sin^2\alpha+2bc\sin^3\alpha. $$ Substituting there $\sin\alpha=a/x$ gives then the requested equality.