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Expanding the squared Euclidean distances givesThus one may takeThe normal is parallel to $\overrightarrow{AB}$, so $L_{AB}$ is its perpendicular bisector. If $\mathbf{x}$ lies on both $L_{AB}$ and $L_{BC}$, then $|\mathbf{x}-\mathbf{a}|=|\mathbf{x}-\mathbf{b}|=|\mathbf{x}-\mathbf{c}|$, hence it also lies on $L_{CA}$. Geometrically, the three perpendicular bisectors of the noncollinear triangle concur at its circumcenter.