Big Ideas Math Geometry, 2014
BI
Big Ideas Math Geometry, 2014 View details
2. Bisectors of Triangles
Continue to next subchapter

Exercise 18 Page 316

To find the circumcenter, we need equations for the perpendicular bisectors of two sides.

Example Solution:

Practice makes perfect

We will start by drawing an arbitrary obtuse triangle, which is a triangle where one angle is greater than 90^(∘). To make it easier for us, we will place one of the sides on the x-axis.

To find the circumcenter, we need equations for the perpendicular bisectors of at least two sides.

Finding perpendicular bisectors

By the Slopes of Perpendicular Lines Theorem, we know that horizontal and vertical lines are perpendicular. Since BC is a horizontal line, the perpendicular line will be vertical. Also, from the diagram, we can identify the midpoint of BC to (3,0). Therefore, our equation for the perpendicular bisector to BC is x=3.
To find the perpendicular bisector to AB, we have to find it's midpoint. We can do that with the Midpoint Formula.
M(x_1+x_2/2,y_1+y_2/2)
M(0+( -4)/2,0+ 4/2)
Simplify
M(0-4/2,0+4/2)
M(- 4/2,4/2)
M(- 2,2)
When we have the midpoint, we will use a protractor to determine where the perpendicular bisector will run.

Finally, using a ruler, we can draw our perpendicular bisector to AB.

Finding the circumcenter

Where the perpendicular bisectors intersect, we find the triangle's circumcenter.

Note that the perpendicular bisector to the triangle's third side will also intersect at the same point as the other two. We could find it as well, but it's enough finding two of them.

Constructing the circumscribed circle

The circumcenter is equidistant to all of the triangle's vertices. Therefore, by using a compass and setting it's radius to the distance between the circumcenter and an arbitrary vertex on the triangle, we can draw a circle that passes through all three vertices.