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 re: Solution by Richard)

In order for (a+b)/2, for example, to equal or exceed 90 degrees, a+b would have to equal or exceed 180; but 180 degrees is the allocation for the whole triangle, including c.

Posted by Charlie on 2006-07-21 15:25:55