Bisector of an Angle of a Triangle

by

Vicki Tarleton

The following is an exploration with proofs involving any triangle and an angle bisector in the triangle.

We will first look at any given triangle ABC. An angle bisector is constructed at vertex C and meets the opposite side at point D. We will refer to segment BC as a, segment CA as b, segment BD as x, and segment DA as y.

Let's explore the relationship of he ratios of side b to a and of y to x. Click here to go to a GSP exploration.

Notice that the ratio of the adjacent sides of the triangle and the ratio of the segments cut off by the bisector are the same. Let's look at the proof for this. Refer to the figure below for the proof.

Notice in the figure, we have constructed a line EB parallel to segment CD. Since CD is an angle bisector, BCD DCA. Since EB is parallel to CD, then ACD AEB because they

are corresponding angles on parallel lines.

Also, DCB CBE because they are alternate interior angles on parallel lines. Since lines CD and EB are parallel, .

Because CEB CBE, CEB is isosceles, making CE CB.

Now by substitution,

Next we will go on to prove that the bisector of an exterior angle of a triangle divides the opposite side externally into segments that are proportional to the adjacent sides.

Using the point X on extended segment BC, ACX is an external angle to ABC. DC bisects ACX making XCD DCA.

Segment AE is drawn such that it is parallel to CD. Therefore, .

Since EA is parallel to CD, then ACD EAC because they are alternate interior anlges on parallel lines.

Also, DCX AEC because they are corresponding angles on parallel lines.

Since CEA CAE, then EC AC.

By substitution, we have .