Given any triangle, through each vertex draw the external angle bisector at that vertex. Show that the new triangle that has as its vertices the three pairwise intersection points of these is always an acute triangle (all three angles strictly less than 90 degrees).

Extra Credit: Extended to meet the new triangle, the internal angle bisectors of the given triangle are what with respect to the new triangle?

(In reply to Solution by JLo)

"These cannot be more than 90 degrees if this holds for a,b,c." However, this does not necessarily hold for a,b,c since you are given any triangle. There is one more little step required to complete the proof, which I'm sure you will figure out very soon.

At one of its points, each extended internal bisector of the given triangle is perpendicular to a side of the new triangle, but there is also another significant  point where it meets the new triangle. Figuring out what this point is will help to answer your final question, I think.

Posted by Richard on 2006-07-21 14:38:53