Date: Dec 1, 1997 12:41 AM
Author: Michael Keyton
Subject: Re: feet of altitudes and angles
here are two different proofs, a few details missing.
If the excenters of a triangle are connected, its orthocenter is the
incenter of the original triangle. This is easily seen since an excenter
is the point of concurrence of the bisectors of two exterior angles and
the bisector of the other interior angle. Since the bisector of an
exterior angle and its adjacent interior angle are perpendicular, the
interior angle bisector is perpendicular to the side of the excenter
triangle, thus is an altitude for it. Since this is true for each of the
interior bisectors, the incenter of the original triangle is the
orthocenter of the excenter triangle. Back these triangles down one level,
and the triangle is the excenter triangle for the orthic triangle (the
triangle formed by the feet of the altitudes.
Proof 2: This is a special case of a little more general situation.
Take a triangle TRI; Draw an altitude from T to a point N on line RI; pick
a point on it, call it P; draw rays from I and R through P to intersect
the sides of TRI at A and W respectively. Thus, P becomes a point of
concurrency for AI, RW and TN, where one of the three is an altitude (TN).
then TN bisects angle INW. (The case for the orthic triangle is a special
case of this theorem.
Proof of the generalized case: Construct a line through T parallel to RI,
extend AI and RW to intersect this line, use similar triangles and the
right angle at N to show angles ANT and WNT are congruent.
I believe I can think of a few more that I have seen from my students over
St. Mark's School of Texas
On Wed, 26 Nov 1997, Annie Fetter wrote:
> My grandmother, who used to be a math teacher, read this problem long ago
> and has always wondered about a proof. (She was never 100% sure that it
> was really true until I modelled the problem in Sketchpad for her. Now she
> is a believer.) I haven't tried to prove it yet, but told her I would
> throw it out to the wolves and see what people came up with.
> Prove that the angles formed by connecting the feet of the altitudes of a
> triangle are themselves bisected by the altitudes.