QuadrangleSolution

The quadrangle is set as the figure on the left. Assume that the lengths $|ab|$, $|bc|$, $|cd|$, $|ad|$ are $2A$, $2B$, $2C$ and $2D$, respectively. Take the mid-points of the sides $ab$, $bc$, $cd$ and $ad$ as in the figure on the right. Join these points to form a square $efgh$. Denote the intersection points of the segment $eh$ and the diagonal $ac$ as $m$, and that of $ef$ and $bd$ as $n$. Then $|ep|=A$, $|fp|=B$, $|gp|=C$, $|hp|=D$. Let $|em|=x$, $|hm|=y$, $|en|=s$, and $|fn|=t$. Thus we have

(a) $x^2+s^2=\color{red}A^2$, (b) $x^2+t^2=\color{red}B^2$, (c) $y^2+t^2=\color{red}C^2$, (d) $y^2+s^2=\color{red}D^2$, (e) $x+y=s+t$.

The area $S$ of the square $efgh$ satisfies $S^2=(x+y)^2=(s+t)^2=(x+y)(s+t)$. Adding up the equations (a)--(d), we get

(f) $2S-2(xy+st)=(x+y)^2+(s+t)^2-2xy-2st=x^2+y^2+s^2+t^2=\frac T2$,

where $\color{red}{}T=A^2+B^2+C^2+D^2$. Next, subtracting (c) from (a) and use the fact that $x-t=s-y$, we get $$\begin{align*}A^2- B^2 & =(x^2-t^2)+(s^2-y^2)=(x-t)(x+t)+(s-y)(s+y)\\&=(x-t)(x+y+s+t)=(s-y)(x+y+s+t).\end{align*}$$ But we also know that $x+y+s+t=2(x+y)=2\sqrt S$. Therefore, the above identities say $$s-y=x-t=\displaystyle\frac{A^2-C^2}{2\sqrt S}.$$

Similarly, subtracting (d) from (b), we have $$x-s=t-y=\displaystyle\frac{B^2-D^2}{2\sqrt S}.$$

Now, from $$\begin{align*}(x-s)^2+(x-t)^2&+(s-y)^2+(t-y)^2\\ &=2(x^2+y^2+s^2+t^2)-2(x+y)(s+t)\end{align*}$$ we infer that $$\displaystyle\frac{ (A^2-C^2)^2+(B^2-D^2)^2)}{2S}=T-2S.$$

This is a quadratic equation $4S^2-2TS+Z=0$ where $\color{red}{}Z=(A^2-C^2)^2+(B^2-D^2)^2$. Solving this quadratic equation to get $S=\frac14\big(T\pm\sqrt{T^2-4Z}\big)$. Note that (f) says that $4S = T+4(xy+st)$ and $xy+st>0$, we conclude that $\color{red}S=\frac14\big(T+\sqrt{T^2-4Z}\big)$. Now, the quadrangle $abcd$ has twice the area of the square $efgh$, namely $2S$.

Note. Any one of the conditions (a), (b), (c) and (d) can be derived by the rest three.