Circumcenter of Triangle
Colleen Foy
Using GSP, we will explore the circumcenter of triangles. Refer to this GSP file.
First we construct a triangle like so:
Next we construct the midpoints of each segment of the triangle:
Now, construct the perpendicular bisectors of each segment of the triangle by constructing perpendicular lines through the midpoints:
The three perpendicular bisectors are concurrent. This is always the case and the proof is left to the reader. Note that we only need two perpendicular bisectors to determine the point of concurrency for all three lines. Now, we construct the point of concurrency of the three perpendicular bisectors using GSP and label it C.
Now construct a circle with center C and one of the vertices using the circle by center point construction tool.
The circumcenter (C) is the center of the circle that passes through all three of the triangle's vertices.
Proof: By definition of perpendicular bisector, C is equidistant from Q and R because it lies on the perpendicular bisector of segment QR. Similarly, C is also equidistant from R and P, and equidistant from P and Q. Thus, C is equidistant to P,Q, and R and P, Q, and R all lie on a circle centered at C.
Note that the above image is for an acute triangle. The circumcenter is inside the circle when the triangle is acute.
Proof:
What happens when the triangle is obtuse?
Notice that when the triangle is obtuse, the circumcenter moves to the outside of the triangle. What happens when the triangle is a right triangle? Note that a right triangle was constructed in GSP then the circumcenter was constructed using the same steps as previously.
It appears that in a right triangle, the circumcenter is the midpoint of the diameter of the circle. Can we prove that the triangle is always a right triangle when the circumcenter lies on the hypotenuse of the triangle?
Proof: Suppose we have a triangle with vertices A, B, and D where A, B, and D all lie on a circle.
Note that DB is the diameter of the circle and C is the circumcenter (we constructed it). Thus,
because they are both radii of the circle. Now construct line segment CA.
Note that
since they are all radii of the circle. Thus, triangle DAC and triangle CAB are both isosceles triangles. Thus, m<ADC = m<DAC and m<ABC = m<CAB by definition of isosceles triangles. Let <ADC and <DAC be called x and let <ABC and <CAB be called y. Note that m<ACB = 2x and m<DCA = 2y because the sum of the angles in a triangle is equal to 180 degrees. Also, note that arc AB + arc AD = 180 because the two arcs make up a semicircle. Thus, m<DCA + m<ACB = 180 since they intercept the arcs AD and AB respectively. By substitution we have that 2y + 2x = 180. Hence x + y = 90. Therefore, triangle ADB is a right triangle.
