The Circumcenter (C) of a triangle is the point of concurrency in the plane equidistant from the three vertices of the triangle. Since a point equidistant from two points lies on the perpendicular bisector from two points lies on the perpendicular bisector of the segment determined by the two points, C is on the perependicular bisector of each side of the triangle. Note: the Circumcenter can be outside the triangle.

 

Prove that the perpendicular bisectors to the sides of the triangle are concurrent.

Given triangle ABC, construct a perpendicular line 'd' bisecting segment AC. Let M be the midpoint of AC.

Construct some point K on the 'd'.

Construct line segments AK and KC.

Prove AK = KC

AM = CM (definition of midpoint)

KM = KM (reflective property)

Angle AMK = Angle CMK (right angles)

Triangle AMK = Triangle CMK (SAS)

AK = CK. QED.

 

Construct a perpendicular line 'g' bisecting BC. Let N be the midpoint of BC.

We know that the perpendicular bisectors 'd' and 'g' intersect because AC and BC are not parallel.

Construct point K so it is the point of intersection of 'd' and 'g'.

Prove BK = CK

BN = CN (definition of midpoint)

KN = KN (reflective property)

Angle BNK = Angle CNK (right angles)

Triangle BNK = Triangle CNK (SAS)

BK = CK. QED.

 

Notice that AK = CK = BK.

It follows that the perpendicular bisector of AB, labeled 'e', will also intersect 'd' and 'g' at point K.

Therefore, the three perpendicular bisectors of triangle ABC are concurrent, as desired. QED.

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