Below we find a triangle ABC in the euclidean plane. The red lines note the perpendicular bisectors for each side of the ABC. Notice that the perpendicular bisectors meet at a single point. As the vertices of ABC are moved about, the intersection is not always necessarily within the triangle.
The more that we move ABC around, we begin to realize that the only way to have the intersection of the perpendicular bisectors not be in within the triangle is move a point such that ABC are colinear. This will, however, violate the fact that a triangle consists of three vertices that are noncolinear.
We can also see that the distances from DA, DB, and DC are equidistant (shown by variables g, h, and i on the left).

Theorem: If a point D is the intersection of the perpendicular bisectors of the sides of triangle ABC, then DA = DB = DC.
Proof: Assume D is an element of the perpendicular bisectors of the sides of ABC. Show DA = DB = DC.
1) Consider the triangle ADB. The perpendicular bisector of ADB splits ADB into ADE and BDE.
BE =AE because DE is the perpendicular bisector of ADB.
Angle(BED) = Angle(AED), for the same reason.
DE = DE
By SAS, ADE is congruent to BDE.
DA = BD, because of corresponding parts of congruent triangles.
2) Consider the triangle ADC. For similar reasoning above, DC = DA.
3) Similar logic for BDC. DB = DC.
Thus, DA = DB = DC.