The following is an exploration
with proofs involving any triangle and an angle bisector in the
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.
CBE because they
are alternate interior angles on parallel lines. Since lines CD
and EB are parallel, .
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
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.
AEC because they
are corresponding angles on parallel lines.
CAE, then EC AC.
By substitution, we have .
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