Proposition 1: The three perpendicular bisectors of any triangle are concurrent.
Proof: We will prove this by proving that it is true for 
ABC, first when ABC is acute, then when it is obtuse, then when it is right.
Case 1-ABC is acute: We will begin by constructing the perpendicular bisectors of sides AB and BC as such:
Because ABC is acute, they will always intersect inside the triangle. We will call their point of intersection R, and the midpoints of the sides of the triangle D, E, and F. We can now construct segments from R to each of A, B, and C, as such:
Now, lets figure some things out:
1. |BE| = |EC| by definition of midpoint.
2. |ER| = |ER| by symmetry.
3. 
BER = CER = 90 degrees by definition of perpendicular.
4. Therefore, BER = CER by SAS. (1, 2, & 3)
5. So |BR| = |CR| by definition of congruence.
Similarly:
6. |AD| = |DB| by definition of midpoint.
7. |DR| = |DR| by symmetry.
8. BDR = ADR = 90 degrees by definition of perpendicular.
9. Therefore, BDR = ADR by SAS. (6, 7, & 8)
10. So |AR| = |BR| by definition of congruence.
11. So |AR| = |CR| by the transitive property. (5 & 10) (Thus |AR| = |BR| = |CR|).
In order to finish the proof, we need to show that FR is perpendicular to AC because we already know F is the midpoint of AC.
12. |CF| = |FA| by definition of midpoint.
13. |FR| = |FR| by symmetry
14. So CFR = AFR by SSS. (11, 12, & 13)
15. So CFR = AFR by definition of congruence.
16. CFA = 180 degrees by definition of straight line
17. CFR + AFR = CFA
18. So CFR = AFR = 90 degrees. (15, 16, & 17)
Therefore, when ABC is acute, the perpendicular bisectors of its sides are concurrent.
Case 2-ABC is obtuse: This proof is the same as the acute case, except the point of concurrency will lie on the outside of ABC. We will begin with a picture:
From here, the proof is exactly the same as Case 1. We use congruent triangles to prove that |BR| = |CR| and |AR| = |BR|. This tells us that |AR| = |BR| = |CR|, which we use to prove CFR = AFR = 90 degrees in the same manner as before. Therefore, when ABC is obtuse, the perpendicular bisectors of its sides are concurrent.
Case 3-ABC is right: We will begin with a picture:
We will set this case up where AB and BC are the legs of the right triangle. As you can see from the picture, the point of intersection of the perpendicular bisectors of AB and BC intersect at point R, and R = E, the midpoint of AC. This is because DR is parallel to BC and ER is parallel to AB. This allows us to see that REC is similar to ABC with a 1 to 2 side ratio, and ADF is similar to ABC with a 1 to 2 side ratio. This 1 to 2 ratios come from the fact that 2|EC| = |BC| and 2|AD| = |AB|
Once we know that these triangles are similar in a 1 to 2 side ratio, we see that ER intersects AC exactly halfway vertically, and DR intersects AC exactly halfway horizontally. Since DBER is a rectangle, both ER and DR must intersect AC at point F, so F = R. Since F = R, the perpendicular bisector of AC must go through point F, so when ABC is right, the perpendicular bisectors of its sides are concurrent.
Corollary of Case 3: |AR| = |BR| = |CR|.
R = F, and |AF| = |CF| by definition of midpoint, so |AR| = |CR|.
|BE| = |CE| by definition of midpoint, BER = CER = 90 degrees by definition of perpendicular, and |ER| = |ER| by symmetry, so BER is congruent to CER by SAS.
Therefore, |BR| = |CR| by definition of congruence, which means |AR| = |BR| = |CR|.
Proposition 2: The intersection of the three perpendicular bisectors of any triangle is the center of the circle that goes through the three vertices of that triangle (the circumcircle).
Proof: In the proof of cases 1 and 2, and in the corollary of case 3, we showed that the point in question (R) is equidistant from all three vertices when we showed that |AR| = |BR| = |CR|. Since the three vertices of a triangle are never linear, and exactly one circle goes through three nonlinear points in the Euclidean Plane, we know that R is the center of the circumcircle.