Assignment #8

 

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) = d(JO)

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 color.

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 side

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 intercepted arc.

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.

Since 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 that

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 stated earlier.....

(The 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]

 

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