7
$\begingroup$

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?

$\endgroup$
  • 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. $\endgroup$ – André Nicolas 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. $\endgroup$ – André Nicolas Nov 23 '13 at 17:22
1
$\begingroup$

This is correct subject to the comments.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.