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?
Let the original triangle be ABC and the new
triangle be A*B*C* where X* lies on the internal
bisector of angle X. The internal and external
bisectors of an angle of ABC are perpendicular.
A*, B*, and C* are the excenters of ABC. ABC is
the orthic triangle of A*B*C*. The internal angle
bisectors of ABC are the altitudes of A*B*C*. Let
I be the incenter of ABC ( it is also the
orthocenter of A*B*C*). Let
m(<A) = 2x
m(<B) = 2y 0 < x,y,z < 90
m(<C) = 2z
IBA*C is a cyclic quadrilateral.
m(<BA*C) = 180 - m(<BIC)
= 180 - ( m(<BIA*) + m(<CIA*) )
= 180 - ( ( m(IAB) + m(IBA) ) +
( ( m(IAC) + m(ICA) ) )
= 180 - ( x+y+z ) - x
= 90 - x < 90
A similar argument holds for m(<CB*A) and
m(<AC*B).
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Posted by Bractals
on 2006-07-21 09:47:37 |
Solution
