Let D XYZ be a triangle. Let line l be the perpendicular bisector of side YZ. Let line m be the perpendicular bisector of side XZ. Let point C be the point of intersection of l and m.
Since C lies on the perpendicular bisector of YZ, then C is equidistant from the endpoints Y and Z. (Perpendicular Bisector Theorem)
Therefore, CY = CZ. (Definition of Equidistant)
Since C lies on the perpendicular bisector of XZ, then C is equidistant from the endpoints X and Z. (Perpendicular Bisector Theorem)
Therefore, CX = CZ. (Definition of Equidistant). CZ = CX. (Symmetric Property)
Since CY = CZ and CZ = CX, then CY = CX. (Transitive Property).
Therefore, C is equidistant from X and Y, the endpoints of side XY. Since C is equidistant from the endpoints of side XY, then C must also lie on the perpendicular bisector of side XY. (Converse of Perpendicular Bisector Theorem). Therefore, the three perpendicular bisectors of D XYZ are concurrent at point C equidistant from the vertices.
This point is called the circumcenter. The circumcenter is the center of the circumscribed
circle of a triangle. For an acute triangle, the circumcenter
lies inside the triangle.
Let D XYZ be an obtuse triangle with < XYZ as an obtuse angle. All the conditions in the proof for an acute triangle hold for an obtuse triangle. For an obtuse triangle, the circumcenter C lies outside the triangle.
Let D XYZ be a right triangle with < XYZ
as a right angle. All the conditions in the proof for an acute
triangle hold for a right triangle. Additionally, note that since
side XY and the perpendicular bisector l are
both perpendicular to side YZ, then l
is parallel to side XY. Then l passes
through the midpoint of side XZ. (Midsegment Theorem)
Since both l and m intersect side XZ at the midpoint E, then the circumcenter (C) is in fact the midpoint of the side opposite the right angle in a right triangle.
Other concurrency theorems
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