Bisectors of Triangles

As you well know by now, being able to deduce key information from a limited set
of facts is the basis of geometry.
An important type of segment, ray, or line that can help us prove congruence is
called an angle bisector. Understanding what angle bisectors are and how
they affect triangle relationships is crucial as we continue our study of geometry.
Let’s investigate different types of bisectors and the theorems that accompany them.

Perpendicular Bisector

A perpendicular bisector is a special kind of segment, ray, or line that

(1) intersects a given segment at a 90° angle, and

(2) passes through the given segment’s midpoint.

Segment CD is the perpendicular bisector to segment AB.

We derive two important theorems from the characteristics of perpendicular bisectors.
We can use these theorems in our two-column geometric proofs, or we can just use
them to help us in geometric computations.

Perpendicular Bisector Theorem

If a point lies on the perpendicular bisector of a segment, then it is equidistant
from the endpoints of the segment.

Converse also true: If a point is equidistant from the endpoints of a segment, then
it lies on the perpendicular bisector of the segment.

These theorems essentially just show that there exist a locus of points (which form
the perpendicular bisector) that are equidistant from the endpoints of a given segment
which meet at the midpoint of the segment at a right angle. An illustration of this
concept is shown below.

Points E, F, G, and H (along with an infinite amount of points) are equidistant from
A and B. Together, they form the perpendicular bisector of segment AB.

The perpendicular bisectors of a triangle have a very special property. Let’s investigate
it right now.

Circumcenter Theorem

The perpendicular bisectors of the sides of a triangle intersect at a point called
the circumcenter of the triangle, which is equidistant from the vertices
of the triangle.

Point G is the circumcenter of ?ABC.

Angle Bisectors

Now, we will study a geometric concept that will help us prove congruence between
two angles. Any segment, ray, or line that divides an angle into two congruent angles
is called an angle bisector.

We will use the following angle bisector theorems to derive important information
from relatively simple geometric figures.

Angle Bisector Theorem

If a point lies on the bisector of an angle, then it is equidistant from the sides
of the angle.

Converse also true: If a point in the interior of an angle is equidistant from the
sides of the angle, then it lies on the bisector of the angle.

The points along ray AD are equidistant from either side of the angle. Together,
they form a line that is the angle bisector.

Similar to the perpendicular bisectors of a triangle, there is a common point at
which the angle bisectors of a triangle meet. Let’s look at the corresponding theorem.

Incenter Theorem

The angle bisectors of a triangle intersect at a point called the incenter
of the triangle, which is equidistant from the sides of the triangle.

Point G is the incenter of ?ABC.

Summary

While similar in many respects, it will be important to distinguish between perpendicular
bisectors and angle bisectors. We use perpendicular bisectors to create a right
angle at the midpoint of a segment. Any point on the perpendicular bisector is equidistant
from the endpoints of the given segment. The point at which the perpendicular bisectors
of a triangle meet, or the circumcenter, is equidistant from the vertices of the
triangle.

On the other hand, angle bisectors simply split one angle into two congruent angles.
Points on angle bisectors are equidistant from the sides of the given angle. We
also note that the points at which angle bisectors meet, or the incenter of a triangle,
is equidistant from the sides of the triangle.

Let’s work on some exercises that will allow us to put what we’ve learned about
perpendicular bisectors and angle bisectors to practice.

Exercise 1

BC is the perpendicular bisector of AD. Find the value
of x.

Solution:

The most important fact to notice is that BC is the perpendicular
bisector of AD because, although it is just one statement, we can
derive much information about the figure from it. The fact that it is a perpendicular
bisector implies that segment DB is equal to segment AB
since it passes through the midpoint of segment AD. Therefore, we
have

Subtracting 7x from both sides of the equation yields

So, we have x=6.

Exercise 2

N is the circumcenter of ?RAK. Find the values of
x and y.

Solution:

We are aware that segments RM, XS, and YE
are all perpendicular bisectors since they meet at N, the circumcenter
of ?RAK. Thus, in order to solve for x, we can set segments
RE and AE equal to each other since E
is the midpoint of segment RA. We have

In order to solve for y, we have to use the information given by the
Circumcenter Theorem. This theorem states that the circumcenter is equidistant
from the vertices of the triangle. Thus, we know that RN=AN=KN. For
this part of the problem, we only need to solve for y with AN=KN.
We have

So, we have x=4 and y=3.

Exercise 3

Find the value of x.

Solution:

The illustration shows that points A and B are equidistant
from point L. By the converse of the Angle Bisector Theorem,
we know that L must lie on the angle bisector of ?AYB.
This means that ?AYL=?BYL, so we can solve for x as
shown below:

So, our answer is x=4.

Exercise 4

QS is the angle bisector of ?PQR. Find the value of
x.

From the information we’ve been given, we know that ?PQS is congruent
to ?SQR because QS bisects the whole angle, ?PQR.
We have been given the measure of the whole angle and the measure of ?SQR,
which is half of the entire angle (since the angle has been bisected). Therefore,
we have

Thus, we get x=9. If we were to plug in 9 for x,
we would see that, indeed, ?SQR is half of ?PQR.