Let $ABCD$ be a circumscribed quadrilateral with center $O$. Assume the incenters of $\vartriangle AOC, \vartriangle BOD$ are $I_1, I_2$, respectively. If circumcircles of $\vartriangle AI_1C$ and $\vartriangle BI_2D$ intersect at $X$, prove the following identity: $(AB \cdot CX \cdot DX)^2 + (CD\cdot AX \cdot BX)^2 = (AD\cdot BX \cdot CX)^2 + (BC \cdot AX \cdot DX)^2$