Its a little similar to G2 in imosl2011

I'll provide an outline of what I believe works. Fix $\triangle DAB$ and move point $C$ uniformly on a fixed line $\ell$. Notice that in terms of distance $d$ from a reference point $O$ on $\ell$, the given expression is a cubic polynomial. To see it is identically zero, we need to find four points on $\ell$ for which it is so. Take three of them to be the intersections with sidelines of $\triangle DAB$ and one to be the intersection with circumcircle of $DAB$. Except for a few degenerate cases, our argument accounts for all lines that intersect the said circle. For any generic line, we can find four of the previous lines that meet it at distinct points. Hence, the result holds for any choice of $C$ in the plane. $\blacksquare$

As MNZ2000 pointed out, this problem looks a lot like 2011 Shortlist G2. Indeed, the same computational approach with signed lengths works. We will instead prove $\frac{W_a+W_b+W_c+W_d}{[ABCD]} = 0$, where $[\mathcal P]$ denotes the (unsigned) area of the polygon $\mathcal P$. Suppose $AC\cap BD = P$, and let $PA = a, PB = b, PC = c, PD = d$; assume $a,b>0$ and $c,d<0$; otherwise we may direct lengths in the opposite directions as necessary. Let $(BCD)$ intersect $AC$ again at $A'$; we have, if $PA' = w$, then $wc=bd$. Therefore, $PA' = \frac{bd}{c}$. Hence, $A'A = PA-PA' = a-\frac{bd}{c}$, so $AA'\cdot AC = \frac{-a+c}{c}\cdot(bd-ac)$. On the other hand, $[BCD]/[ABCD] = \frac{-c}{a-c}= \frac{c}{-a+c}$. So, $$\frac{W_a}{[ABCD]} = bd-ac.$$Our similar expressions for $W_b,W_c,W_d$ yield the conclusion.

I'll post a more clear solution. The previous one had the right idea but lacks rigour (and even that it's not a cubic). I hope it works, but if not, please let me know the flaw MNZ2000 wrote: Let $ABCD$ be a convex quadrilateral and $W_a$ the product of the power of $A$ about circle $BCD$, and the area of triangle $BCD$. Define $W_b,W_c,W_d$ similarly. Prove that $$W_a+W_b+W_c+W_d=0.$$ Claim: Let $ABC$ be a triangle and $\ell$ be a line. Point $P$ varies uniformly over line $\ell$, say, at distance $p \in \mathbb{R}$ from a fixed point; then the power of point $A$ in circle $(BPC)$ is given by a rational function $\frac{f(x)}{g(x)}$ in $p$ with $\text{deg} \, f \le 2$ and $\text{deg} \, g \le 1$. (Proof) Notice that the equation of $(BPC)$ is $$ux(x+y+z)-a^2yz-b^2zx-c^2xy=0.$$Plug $A=(1,0,0)$ to see that it has power equal to $u$. Plug $P=(x', y', z')$ to see that $u$ has the desired form since $x', y', z'$ are linear functions of $p$ with $x'+y'+z'=1$. $\blacksquare$ Fix $\triangle DAB$ and a line $\ell$. Move point $C$ on line $\ell$ with constant velocity. From a reference point $O$ on $\ell$, let $C$ be at distance $t \in \mathbb{R}$. Let $[BCD]=h_A(t)$ and define $h_B, h_C, h_D$ analogously. Note that $h$'s are linear functions of $t$ with only $h_C$ as constant. Also, $$p(A, (BCD))=\frac{f_A(t)}{g_A(t)}$$where $f_A, g_A$ are polynomials of degree $\le 2$ and $\le 1$ respectively. The crucial observation is that $f_A, f_B, f_C, f_D$ all vanish when $ABCD$ is cyclic; hence they all have a quadratic common factor $q \in \mathbb{R}[X]$. Also, $g_C$ is a constant polynomial, and for $x \in \{D, A, B\}$ we see that both $g_x, h_x$ vanish when $C$ is on the line $yz$ with $\{x, y, z\}=\{D, A, B\}$. Hence, $g_x, h_x$ have a common linear factor. We conclude that $$\sum_{\text{cyc}} W_a=\sum_{\text{cyc}} \frac{f_A(t)h_A(t)}{g_A(t)}=q(t) \cdot r(t),$$where $r$ is a constant polynomial. To conclude, the previous analysis implies that it suffices to show the result when $ABCD$ is a non-cyclic, non-degenerate quadrilateral, for any one point $C$ in the plane. Pick $C$ such that $ABCD$ is a parallelogram. It is immediate that $\sum_{\text{cyc}} W_a=0$ in this case; hence the result is true for all configurations, just as we desired. $\blacksquare$

Using reim and thales and ratio of areas for triangles with the same altitude, we get that all the $X_A$'s have the same magnitude but 2 of them are negative and others are positive, so done. Note: The original problem had $X_A$'s and not $W_A$'s .