Proving Angle Bisector Problem in Non-Isosceles Triangle

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The discussion centers on proving that the interior angle bisectors of two angles and the exterior bisector of the third angle in a non-isosceles triangle intersect the sides at three collinear points. A participant expresses difficulty in progressing with the proof, noting that while the interior bisectors create two collinear points, this does not aid in the overall proof. Suggestions include breaking the triangle into small rectangles to analyze the centers of gravity. The conversation emphasizes the geometric relationships within the triangle to establish the collinearity of the points. The goal is to achieve a clear proof of this geometric property.
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Prove that the interior bisectors of two of the angles of a non-isosceles triangle and the exterior bisector of the third angle meet the sides of the triangle in three collinear points.


I hope this is posted in the right area because it is concerning geometry!

I've been trying at this for a few days and can't make any progress. I understand that the two points formed from the interior bisectors are collinear, but that really doesn't help because any points two points are collinear. So any help is appreciated.
 
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The rought and ready answer is to split the triangle up into very tiny rectangles, the centra of gravity (which is what your asking for really) will be in the middle of each of these very thin rectangles.

Do this for all three sides to obtain the answer.

Mat
 
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