Lesson 10 Centers of a Triangle Practice Understanding

Learning Focus

Examine properties of the medians, angle bisectors, and perpendicular bisectors of the sides of triangles.

Construct the center of a circle that will pass through all three vertices of a triangle and the center of a circle that can be drawn in a circle so that it touches all three sides.

Construct the “balancing point” of a triangle.

Given any triangle, can a circle be drawn that passes through all three vertex points of the triangle? Can a circle be drawn that is tangent to all three sides of the triangle? If so, how do we find the center of such circles?

Open Up the Math: Launch, Explore, Discuss

Kolton, Kevin, and Kara have been asked by their fathers to help them solve some interesting geometry problems.

Kolton’s Problem

Kolton’s father installs sprinkling systems for farmers. The systems he installs are called “Center Pivot Irrigation Systems” since the sprinklers are on a long pipe that rotates on wheels around a center point, watering a circular region of crops. You may have seen such “crop circles” from an airplane.

Sometimes Kolton’s father has to install sprinkler systems on triangular-shaped pieces of land. He wants to be able to locate the “pivot point” in the triangular field so the circle being watered will touch each of the three fences that form the boundaries of the field. He has asked for Kolton’s help with this problem, since Kolton is currently studying geometry in high school.

Kara’s Problem

Kara’s father installs cell towers. Since phone signals bounce from tower to tower, they have to be carefully located. Sometimes Kara’s father needs to locate a new tower so that it is equidistant from three existing towers. He thinks of the three towers that are already in place as the vertices of a triangle, and he needs to be able to find a point in this triangle where he might locate the new tower so that it is equidistant from the other three. He has asked Kara to help him with this problem since she is also studying geometry in school.

Kevin’s Problem

Kevin’s father is an artist and has been commissioned by the city to build an art project in the park. His proposal consists of several large pyramids with different shaped triangles balanced on the vertex points of the pyramids. Kevin’s father needs to be able to find the point inside of a triangle that he calls “the balancing point.” He has asked Kevin to use his knowledge of geometry to help him solve this problem.

Kolton, Kevin, and Kara’s geometry teacher has suggested they try locating points in the interior of triangles where the three medians, the three angle bisectors, or the three perpendicular bisectors of the sides intersect.

1.

Try out the experiment suggested by the students’ geometry teacher. Which set of line segments seem to locate a point in the triangle that best meets the needs of each of their fathers? (Your teacher will provide some tools to assist you in this exploration.)

Kolton, Kevin, and Kara have noticed something interesting about these sets of line segments. To their surprise, they notice that all three medians of a triangle intersect at a common point. Likewise, the three altitudes also intersect at a common point. So do the three angle bisectors, and the three perpendicular bisectors of the sides. They think their fathers will find this interesting, but they want to make sure these observations are true for all triangles, not just for the ones they have been experimenting on. The diagrams and notes given suggest how each is thinking about the proof they want to show his or her father. Use these notes and diagrams to write a convincing proof.

Kolton’s Notes

What I did to create this diagram:

I constructed the angle bisectors of angle $A C B$ and angle $A B C$ . They intersected at point $P$ . So I wouldn’t get confused by so many lines in my diagram, I erased the rays that formed the angle bisectors past their point of intersection. I then drew ray $A H$ through point $P$ , the point of intersection of the two angle bisectors.

My question is, “Does this ray bisect angle $C A B$ ?” While I was thinking about this question, I noticed that I had created three smaller triangles in the interior of the original triangle. I constructed the altitudes of these three triangles (they are drawn as dotted lines). When I added the dotted lines, I started seeing kites in my picture. I’m wondering if thinking about the smaller triangles or the kites might help me prove that ray $A H$ bisects angle $C A B$ .

2.

Complete Kolton’s argument:

Kara’s Notes

What I did to create this diagram:

I constructed the perpendicular bisectors of side $A C$ and side $B C$ . They intersected at point $P$ . So I wouldn’t get confused by so many lines in my diagram, I erased the rays that formed the perpendicular bisectors past their point of intersection. I then constructed a line perpendicular to side $A B$ through point $P$ , the point of intersection of the two perpendicular bisectors. I named the point where this line intersected side $A B$ point $H$ .

My question is, “Does this perpendicular line also bisect side $A B$ ?” While I was thinking about this question, I noticed I had created some quadrilaterals in the interior of the original triangle. Since quadrilaterals in general don’t have a lot of interesting properties, I decided to make some triangles by dotting in line segments drawn from $P$ to each of the vertices of the original triangle. I’m wondering if thinking about these smaller triangles might help me prove that line $P H$ bisects side $A B$ .

3.

Complete Kara’s argument:

Ready for More?

Kevin’s Notes

What I did to create this diagram:

Point $M$ is the midpoint of side $A C$ , and point $N$ is the midpoint of side $C B$ . Therefore, $\overset{―}{A N}$ and $\overset{―}{B M}$ are medians of the triangle. I then drew ray $C F$ through point $P$ , the intersection of the two medians. My question is, “Does this ray contain the third median?” So, I need to find a way to answer that question.

As I was thinking about this, I thought I could visualize a parallelogram with its diagonals, so I drew line $G B$ to be parallel to median $A N$ , and then connected vertex $A$ to point $G$ on the ray. Quadrilateral $A G B P$ looks like a parallelogram, but I’m not so sure. And I am wondering if that will help me with my question about the third median. What do you think?

Complete Kevin’s argument:

[Hint: Kevin’s proof uses similar triangles. Corresponding angles in similar triangles are congruent. Can you find a pair of triangles that you know are similar? Can you find a pair of triangles that you can prove are similar?] You will work with similar triangles in Unit 4, so you may want to revisit this problem as part of your work in that unit.

Takeaways

Given any triangle, a circle can be drawn that passes through all three vertex points of the triangle.

To locate the center of the circle,

Given any triangle, a circle can be drawn that is tangent to all three sides of the triangle.

To locate the center of the circle,

Given any triangle, a balancing point can be found for the triangle.

To locate the balancing point,

Adding Notation, Vocabulary, and Conventions

Three or more lines or line segments that meet at a single point are said to be , since this requires more constraints and is more unique than when two lines intersect at a single point. The point where the set of three or more lines or line segments meet is called the .

If this point is formed by the three angle bisectors of a triangle, it is also called the , since it locates the center of a circle that can be in the triangle, that is, a circle that is tangent to (just touches) all three sides of the triangle.

If this point is formed by the perpendicular bisectors of the three sides of a triangle it is called the , since it locates the center of a circle that can be about the triangle, that is, a circle that passes through all three vertices of the triangle.

The balancing point of a triangle is called the .

Vocabulary

Lesson Summary

In this lesson, we found that the three medians of a triangle are concurrent, which means that they all meet at the same point on the interior of a triangle. We also found that the three angle bisectors of a triangle are concurrent, as well as the three perpendicular bisectors of the sides. These points of concurrency are called centers of a triangle, since they locate interesting points in the interior or exterior of the triangle, such as the balancing point or the centers of circles that can be drawn to pass through all three vertices of the triangle or touch all three sides.

Retrieval

Use the diagram of the parallelogram to find the measures listed.

1.

$D E$

2.

$m ∠ C G F$

3.

$m ∠ C E F$

4.

$D G$