Let´s consider a general triangle ABC. Let´s draw two angle bisectors from vertices A and B.

It is obvious that these two angle bisectors intersect at a single point X. Since X lies on the angle bisector from A its distance from side b is the same as its distance from side c. As X also lies on the angle bisector from B, its distance from side c is the same as its distance from side a.

By transitivity the distance from the point X to the side b has to be the same as the distance from X to the side a. Therefore the angle bisector from C (which is the set of points equidistant from b and a) has to go through X.

It seems as a very simple argument. Is it correct?

  • 2
    $\begingroup$ It is correct, if the theorem that $P$ is on the angle bisector if and only if it is equidistant from the two intersecting lines has already been proved. $\endgroup$ Nov 23 '13 at 17:16
  • 1
    $\begingroup$ It has been proved. Not without some struggle, though. $\endgroup$
    – Adam
    Nov 23 '13 at 17:19
  • $\begingroup$ It comes down to two congruence of triangles arguments. After that, your argument for the concurrence of the three angle bisectors is good. $\endgroup$ Nov 23 '13 at 17:22

This is correct subject to the comments.


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