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Given a convex quadrilateral $ABCD$ that isn't cyclic, prove that the circumcenters of $\triangle{ABC},\triangle{ABD},\triangle{BCD},$ and $\triangle{ADC}$ aren't collinear.

Let $A_1$ denote the circumcenter of $\triangle BCD$. Define $B_1, C_1,D_1$ in a corresponding way.

I don't see how to solve this, but I would try proof by contradiction. That is, assume, without loss of generality, that $A_1,B_1,C_1,D_1$ are collinear. We then would have to show a contradiction. I would say that since the three radical axes are parallel, there does not exist a point with equal power to the four circles. I am not sure how to continue from here.

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Take any two of those 4 circles with centers $O,O'$; then they meet at exactly two points $X,Y\in\{A,B,C,D\}$ (by the condition that $ABCD$ is not cyclic). Thus $XY\perp OO'$. So even if three of the 4 centres are collinear, it must be that the corresponding segments $XY$ for the three pairs among them are parallel. However, it's not possible for three segments with vertices among $\{A,B,C,D\}$ to be parallel because some two will have a common point.

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As $ABCD$ is not cyclic, none of the four centres coincide. The circumcentres of $\Delta ABC$ and $\Delta ABD$ are both on the perpendicular bisector of $AB$. The circumcentres of $\Delta ABC$ and $\Delta BCD$ are both on the perpendicular bisector of $BC$. We conlcude that these two perpendicular bisectors coincide. But then $A=C$ (and $ABCD$ is cyclic) as both are obtained by reflecting $B$ at these bisectors.

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