The sum is 1, always 1

Note that if a triangle' sides are x, y, z, and if its area is t, then $t=\dfrac{abc}{4R}\Leftrightarrow 4tr=abc$

So $a\times b\times c\times d=\dfrac{4t_{ABC}}{AC}\times\dfrac{4t_{CDA}}{AC}=\dfrac{16\frac{AC\times x}{2}\times \frac{AC\times y}{2}}{AC^2}=\dfrac{4AC^2xy}{AC^2}=4xy\leq (x+y)^2\leq 4$

The equality will only be true, if $x+y=2$ and $x=y$ , so $x=y=1$ . Now it is clear that the quadrilateral is a square, and its area is $\boxed{2}$ .