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) $x^2+s^2=A^2$, (b) $x^2+t^2=B^2$, (c) $y^2+t^2=C^2$, (d) $y^2+s^2=D^2$, (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) $2S-2(xy+st)=(x+y{)}^2+(s+t{)}^2-2xy-2st=x^2+y^2+s^2+t^2=\frac{T}{2}$,

where $T=A^2+B^2+C^2+D^2$. Next, subtracting (c) from (a) and use the fact that $x-t=s-y$, we get $$\begin{array}{rl}{A}^2-B^2& =(x^2-t^2)+(s^2-y^2)=(x-t)(x+t)+(s-y)(s+y)\\ & =(x-t)(x+y+s+t)=(s-y)(x+y+s+t).\end{array}$$ 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 $$\begin{array}{rl}(x-s{)}^2+(x-t{)}^2& +(s-y{)}^2+(t-y{)}^2\\ & =2(x^2+y^2+s^2+t^2)-2(x+y)(s+t)\end{array}$$ 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 $Z=(A^2-C^2{)}^2+(B^2-D^2{)}^2$. Solving this quadratic equation to get $S=\frac{1}{4}(T\pm \sqrt{T^2-4Z})$. Note that (f) says that $4S=T+4(xy+st)$ and $xy+st>0$, we conclude that $S=\frac{1}{4}(T+\sqrt{T^2-4Z})$. 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.