Proposition: The perpendicular bisectors of any triangle meet in a single point [which is equidistant from the vertices]

1/2 *v + n = *1/2 *w + m =* *v + *1/2* z - p*

To show*p* is perpendicular to *z* it is enough to show that the inner (dot) product of *p z = *0.

Now*p = v +*1/2 *z* - 1/2 *v - n = *1/2 *v + *1/2 (*w-v*)* - **n* = 1/2 *w -* *n* , so

*p z = *(1/2 *w - **n*) (*w - v*) = 1/2 *w w - wn* - 1/2 *w v + n v. *But since *n* is perpendicular to AB,

nv =0 so we have

*p z = *1/2 *ww - wn* - 1/2 *v w =**w* (1/2 * w - n* - 1/2* v*)

=*w *( 1/2 *w - *(1/2 *v + n*)) =*w *( 1/2 *w - *(1/2 *w + m*))= *w *(-*m*) = 0

because *m *is perpendicular to AC.

From the geometry is is clear that the point O must be equidistant from
A, B, C and thus is the center of the circle that passes through the
points A, B, and C, - the circumcenter of the triangle ABC.

To show

Now

nv =

=