The external bisector of angle A of A triangle ABC meets BC produced at L. And the internal bisector of angle B meets CA at M. If LM meets AB at R. Prove that CR bisects angle C?

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Given:

Triangle ABC

External bisector of angle A meets BC produced at L

Internal bisector of angle B meets CA at M

LM meets AB at R

To prove: CR bisects angle C

Proof:

Lemma 1: The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the adjacent sides.

Proof of Lemma 1: Let ABC be a triangle and AD be its external bisector of angle A. Then, by angle bisector theorem, we have:

BD/DC = AB/AC

By exterior angle theorem, we have:

∠BAC = ∠ABD + ∠ADC

But, ∠ABD = ∠ADC (as AD is an external bisector)

So, ∠BAC = 2∠ABD

Therefore, in triangle ABD, we have:

sin(∠ABD)/BD = sin(∠BAC/2)/AB

Similarly, in triangle ADC, we have:

sin(∠ADC)/DC = sin(∠BAC/2)/AC

Dividing the two equations, we get:

BD/DC = AB/AC

which proves the lemma.

Lemma 2: The internal bisectors of two angles of a triangle intersect at a point which is equidistant from the sides of the third angle.

Proof of Lemma 2: Let ABC be a triangle and AD and BE be its internal bisectors of angles A and B respectively. Then, by angle bisector theorem, we have:

BD/DC = AB/AC ...(1)

and

AE/EC = AB/BC ...(2)

Multiplying (1) and (2), we get:

(BD/DC)*(AE/EC) = 1

By converse of angle bisector theorem, we have:

AD and BE intersect at a point I which is equidistant from sides AC and BC.

Proof of the main result: Since CR is the third transversal of the lines AL and BM, we have:

(BL/LC) * (CA/AM) * (MR/RB) = 1

By Lemma 1, we have:

BL/LC = AB/AC

By Lemma 2, we have:

CA/AM = CB/BM

Therefore, we have:

(AB/AC) * (CB/BM) * (MR/RB) = 1

which implies:

MR/RB = AC/AB

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