Another isosceles triangle problem:  Obviously, if
two medians or two altitudes of a triangle are congruent,
then the triangle is isosceles.  What about if the
lengths of two angle bisectors are congruent? (from the
bisected verticies to the opposite side).  Does this
imply that the triangle is isosceles?  Is there an elementary
proof for this?  Thanks!

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Request for Question Clarification by mathtalk-ga on 04 Nov 2003 08:19 PST

Hi, whack-ga:

I hope you saw the Comment by racecar-ga.  It's a nice reference, and
I think you'll be pleased with it.  Some of the difficulty is "hidden"
in the assumed result that the three angle bisectors are concurrent.

Take a bow (and a victory lap?), racecar-ga.

regards, mathtalk-ga

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Clarification of Question by whack-ga on 04 Nov 2003 11:16 PST

mathtalk-ga,

I agree, it's a nice proof--but is does assume that the
angle bisectors all intersect, which is not an obvious result--
still, I think it's easier to follow than a trig proof and I could
certainly use it in a Geometry class!  Thanks, racecar-ga!

(I am of course happy to pay for the answer)!

whack-ga

It's true, but there is no elementary proof. A more involved proof can be found at
http://www.mathpages.com/home/kmath433.htm

I had read somewhere in a bibliography that 
there was a direct Geometric proof that did 
not use trigonometry.  I'm hoping that someone
has seen or heard of it.  Thanks for the link--
I had no idea the proof was so complicated!

whack-ga

A cute proof is given at:

http://rec-puzzles.org/sol.pl/geometry/bisector