### Theorem: Suppose that the internal angle bisectors of triangle ABC are extended to meet the circumcircle of triangle ABC at the points L, M, and N respectively. Then the following relationships hold:

#### The measure of angle L is half the sum of the measures of angles B and C. The measure of angle M is half the sum of the measures of angles A and C. The measure of angle N is half the sum of the measures of angles A and B.

To prove the theorem, we will need the following lemma:

#### Lemma: Let C be a circle with center O, and let the actue angle XYZ be inscribed in circle C. Then the measure of angle XYZ is half the measure of angle XOZ.

When triangle LMN is constructed in the manner described in the statement of the problem, triangle LMN is always acute. Therefore, we can use the lemma in the proof of the theorem, regardless of whether triangle ABC is acute, right, or obtuse.

### Proof of Theorem:

Construct the circumcenter O of triangle ABC. Construct the segments ON and OM.
From the lemma, we have,

#### m<L = 1/2 m<MON.

Since A is between M and N,

So we now have

#### m<L = 1/2 m<MON = 1/2 (m<MOA + m<AON) = 1/2 m<MOA + 1/2 m<AON.

Consider the angle MBA. Since M is on the bisector of angle CBA, the measure of angle MBA is half the measure of angle CBA. Since angle CBA is an angle in a triangle, the measure of angle CBA is less than 180 degrees. Therefore, the measure of angle MBA is less than 90 degrees, so it is acute.

Similarly, the measure of angle ACN is half the measure of angle ACB, so the measure of angle ACN is less than 90 degrees.

Since angles MBA and ACN are both acute, we can apply the lemma to them to get

and

So now we have

#### m<L = 1/2 m<MOA + 1/2 m<AON = m<MBA + m<ACN.

Since M is on the bisector of angle CBA,

Similarly,

Now we have

Similarly,

and

#### m<N = 1/2 (m<A + m<B).

This proof does not require that triangle ABC be acute, so the result holds for acute, right, and obtuse triangles.

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