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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?

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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.

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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
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