Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In triangle ABC let X be the point of tangency of the excircle opposite A with side BC. (A) Prove that the segment AX divides triangle ABC into two triangles, each having the same perimeter. (B) Prove or disprove: The point X must lie on the nine-point circle of triangle ABC.

I really need some help with parts (A) and (B). I think for (B) that it should be proved true because the three excircles and the incenter are all tangent to the nine-point circle by Feuerbach's theorem. Is this correct? How should I prove this using that theorem if that is the right direction to head?

share|cite|improve this question
up vote 1 down vote accepted

Well, for part A) the only thing you need to prove is that $AB+BX+AX=AC+CX+AX.$ If $B_a$ and $C_a$ are the points where excircle touches $AC$ and $AB,$ then $AB_a=AC_a$ and $CB_a=CX,$ $BC_a=BX.$ Now $AB_a=AC+CB_a=AC+CX=AC_a=AB+BC_a=AB+BX$ and the result follows.

Now the point $X$ does not belong to the Euler's circle. Indeed, Euler's circle passes through the midpoint of the side $BC$ as well as through the projection of $A$ on the side $BC.$ Clearly, none of the two points mentioned above coincides with $X$ in general and circle cannot have more than two points of intersection with a straight line.

share|cite|improve this answer
Is there an instance when the incircle, nine-point circle, and the excircle would meet at the Feuerbach point on side BC and then that would actually prove it? – Josh May 9 '13 at 5:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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