Locus equidistant from four points

Problem

Prove that, given four non-coplanar points $A,B,C,D$ in space, there is exactly one point equidistant from all four, and that this point is the centre of the sphere through $A,B,C,D$.

Solution

The locus of points equidistant from two points $A,B$ is the perpendicular bisector plane of segment $AB$ (plane perpendicular to $AB$ at its midpoint). Points equidistant from $A,B,C$ (with $A,B,C$ non-collinear) lie on the intersection of two such planes: a line perpendicular to plane $ABC$ through the circumcentre of triangle $A\overset{\triangle}{B}C$. Imposing equidistance from $D$ as well adds a third bisector plane; since $D$ is not coplanar with $A,B,C$, this plane does not contain the line and meets it in exactly one point $O$. This $O$ is equidistant from $A,B,C,D$ and is therefore the centre of the sphere through the four points. $\boxed{\,O=\text{intersection of the bisector planes}=\text{centre of the sphere through }A,B,C,D\,}$