**Angle Bisectors of aTriangle**

By Nikhat Parveen, UGA.

*Given that the internal angle
bisectors of triangle ABC that are extended to meet the circumcircle at points
L, M, and N, respectively we wish to find the angles of triangle LMN in terms of
the angles A, B, and C. *

We will proceed by drawing a GSP sketch of the given figure in the following steps:

1. Drawing an acute triangle and circumscribing a circle.

2. Constructing the angular bisectors of each angle of a triangle.

3. Marking the intersections of the circumcircle and the angular bisectors (as each angular bisector will intersect the circumcircle) as A, B, and C and L, M, N respectively.

This will result in the figure shown below:

Now that we have the figure we
are interested in finding the angles L, M, and N in terms of A, B, and C. By
noticing at the figure closely we observe that the intersecting points on the
circumscribing circle are the intercepted angles of the circumscribing circle.
Now we have to go back to the basics of geometry, by which I mean referring to
the important theorems in geometry. One of the the theorems relating to the
inscribed angles of a circle is *
"The measure of an inscribed angle is one-half
the measure of its central angle". (For
proof, please click
HERE.)
*

A corollary of this theorem
states that
"when two inscribed angles of a circle intercept
a common arc, then the angles are equal".
(click **
here** to
see a proof).
We will use the corollary mostly in finding a
solution to the problem.

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**Consider the figure below:**

Since ray BM is the angle bisector of angle B, then the measure of angle ABM = 1/2 of the measure of angle B. Since inscribed angles ABM and ALM intercept a common arc (see sketch above), by the corollary, the measure of angle ABM = the measure of angle ALM. By substitution, the measure of angle ALM = 1/2 of the measure of angle B.

Since ray CN is the angle bisector of angle C, (as shown in the figure below) then the measure of angle ACN = 1/2 of the measure of angle C. Since inscribed angles ACN and ALN intercept a common arc (see sketch below), by our corollary, the measure of angle ACN = the measure of angle ALN. By substitution, the measure of angle ALN = 1/2 of the measure of angle C.

Therefore by angle addition, the measure of angle L = the sum of the measures of angle ALM and ALN. By substitution, we get the measure of angle L = 1/2 of the measure of angle B + 1/2 of the measure of angle C

**Consider the figure below:**

As we notice, ray AL is the angle bisector of angle A, then the measure of angle BAL = 1/2 of the measure of angle A. Since inscribed angles BAL and BML intercept a common arc (see sketch above), by our corollary, the measure of angle BAL = the measure of angle BML. By substitution, the measure of angle BML = 1/2 of the measure of angle A.

Now ray CN is the angle bisector of angle C, then the measure of angle BCN = 1/2 of the measure of angle C. Since inscribed angles BCN and BMN intercept a common arc (see sketch below), by our corollary, the measure of angle BCN = the measure of angle BMN. By substitution, the measure of angle BMN = 1/2 of the measure of angle C.

Therefore by angle addition, the measure of angle M = the sum of the measures of angle NMB and BML. By substitution, we get the measure of angle M = 1/2 of the measure of angle A + 1/2 of the measure of angle C.

Consider the figure below:

From the figure we see that ray BM is the angle bisector of angle B, then the measure of angle CBM = 1/2 of the measure of angle B. Since inscribed angles CBL and CNM intercept a common arc (see sketch above), by our corollary, the measure of angle CBM = the measure of angle CNM. By substitution, the measure of angle CNM = 1/2 of the measure of angle B.

Since ray AL is the angle bisector of angle A, then the measure of angle CAL = 1/2 of the measure of angle A. Since inscribed angles CAL and CNL intercept a common arc (see sketch below), by our corollary, the measure of angle CAL = the measure of angle CNL. By substitution, the measure of angle CNM = 1/2 of the measure of angle A.

Therefore by angle addition, the measure of angle N = the sum of the measures of angle CNM and CNL. By substitution, we get the measure of angle N = 1/2 of the measure of angle B + 1/2 of the measure of angle A.

In summary we found:

Angle 'L' = measure of 1/2 (angle B + angle C)

Angle 'M' = measure of 1/2 (angle A + angle C)

Angle 'N' = measure of 1/2 (angle A + angle B)

combining we get angle (L + M + N) = measure of angles 1/2( 2A + 2B + 2C)

= measure of angles (A + B + C)

= 180 degrees

= sum of the angles in a triangle ABC

This verifies the measure of angles L, M, and N in terms of angle A, B and C holds true.

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Does our result hold only for acute triangles? Well, we can verify this by considering obtuse angles and right angles.

I investigated it using GSP and found that the conjecture we proved holds true for the obtuse angles as well as with right angles. Below is one such observation. For further manipulation and to check out by self, please click here.

Click HERE for the exploration in GSP

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