Circumcenter Theorem

Assume that ABC is a triangle and DS, ES, and FS are the perpendicular bisectors of the sides of this triangle.

Notice that S is a point on the perpendicular bisector of AB. Therefore, by the Perpendicular Bisector Theorem, S is equidistant from A and B.

Similarly, S is also a point on the perpendicular bisector of BC. Using the Perpendicular Bisector Theorem once again, it can be concluded that S is equidistant from B and C.

By the Transitive Property of Equality, AP is equal to CP. AS= BS BS=CS ⇒ AS=CS This proves that AS, BS, and CS are all equal.

Two-Column Proof

The proof can be summarized in the following two-column table.

Statements	Reasons
1. & ABCis a triangle & DSis a perpendicular bisector ofAB & ESis a perpendicular bisector ofBC & FSis a perpendicular bisector ofAC	1. Given
2. & AS=BS & BS=CS	2. Perpendicular Bisector Theorem
3. AS=CS	3. Transitive Property of Equality