Finding the Angles
of a Triangle by Extended Angle Bisectors
By: Elizabeth Gore
What is the problem?
The internal angle
bisectors of triangle ABC are extended to meet the circumcircle
at points K, L,and J, respectively.
Find the angles of
triangle KLJ in terms of the angles A, B, and C.
First we will take
a look at the structure we will be working with.
Below is the given
triangle, ABC, with its angle bisecotrs marked in green and the
circumcircle drawn around it.
From the angle bisectors of the given
triangle we have drawn a new triangle KLJ.
Our task is to find the angles of KLJ
using the angles of ABC.
We will begin by first constructing
segments that link each vertex with the circumcenter of triangle
ABC labled O.
The claims we have made above are revealed
through a Geometer's Sketch Pad investigation
showing that the perpendicular bisectors
of triangle ABC actually contain the points K, L, and J.
So the claim made above is indeed true.
By the definition of circumcircle of
triangle ABC, we also know that O lies on each of the perpendicular
bisectors, as well.
We know that by the definition of a
circle, since O is the center of our circle, then:
d(AO) = d(KO) = d(BO) = d(LO) = d(CO)
Above we noted that each of our segments
are the same length.
Therefore, within triangle ABC, we
have three isosceles triangles.
Each triangle is highlighted a different
Let's take a qick look at the triangle
theorum that states:
The line drawn from
the vertex of an isosceles triangle to the midpoint of the opposite
bisects the angle at
the vertex and is perpendicular to the opposite side.
This will come in handy!!
So we have that:
OK bisects AB
OL bisects BC
OJ bisects AC
Now that each of our
three triangle has been bisected into two new triangles, we can
begin totalk about their angles
Since OK bisects AB
at the intersection point Y,
m(AOY) = m(YOB)
Using a similar argument,
we get that
m(BOZ) = m(ZOC)
m(COX) = m(XOC)
Since angle measure is additive, we
can see that
m(AOB) = m(AOY) + m(YOB)
m(BOC) = m(BOZ) + m(ZOC)
m(COA) = m(COX) + m(XOA)
To successfully solve
our problem, we must know the following theorum about circles.
The measure of an inscribed
angle is equal to 1/2 the measure of the central angle of its
With that theorum in
mind, we can say that the measure of the inscribed angle BAC
is equal to 1/2 the
measurement of the central angle BOC.
Below I have actually
measured the angles to prove that this theorum works.
Here we see some simple arithmetic.
m(BAC) = 1/2 m(BOC), and m(BOC) = m(BOZ) + m(COZ),
Then m(BAC) must equal
m(BOZ) which equals m(COZ).
A similar argument is made to show
m(ABC) = 1/2m(AOC)
= m(AOX) = m(XOC)
m(ACB) = 1/2m(AOB)
= m(AOY) = m(YOB)
Continuing with this line of reasoning,
we can make a similar argument for triangle KLJ.
We get that:
m(KLJ) = 1/2m(KOJ)
= m(KOA) = m(JOA) = m(YOA) = m(XOA)
m(KJL) = 1/2m(KOL)
= m(KOB) = m(LOB) = m(YOB) = m(ZOB)
m(LKJ) = 1/2m(LOJ)
= m(LOC) = m(JOC) = m(ZOC) = m(XOC)
m(KOJ) = m(KOA) + m(JOA)
= m(YOA) + m(XOA) = m(ACB) + m(CBA)
m(KOL) = m(KOB) + m(BOL)
= m(YOB) + m(ZOB) = m(BAC) + m(ACB)
m(LOJ) = m(LOC) + m(COJ)
= m(ZOC) + m(XOC) = m(CBA) + m(BAC)
Keeping in mind the theorum about circles
measure of an inscribed angle is equal to 1/2 the measure of the
central angle of its intercepted arc.)
we are finally left with three simple
formulas to measure the angles of triangle KLJ.
angle L = 1/2 [angle C + angle B]
angle J = 1/2 [ angle C + angle A]
angle K = 1/2 [ angle A + angle B]