(The converse is easy.)

It depends on what you mean by "elementary." There's a quick proof if you use the two facts that (1) the angles bisectors of a triangle intersect at the incenter, the center of the inscribed circle, and (2) tangents drawn from a point external to a circle are congruent.

(To get you started: Let the triangle be ABC with the angle bisector of A hitting BC at D and the angle bisector of B hitting AC at E. You're assuming AD=BE and you want to show that AC=BC, or equivalently, angle A is congruent to angle B. Draw the inscribed circle, and look at it from the perspective of point C...)

## Comments

As person posing the problem - 1) I withdraw the qualifier "elementary" for the type of proof desired. 2) I'd appreciate knowing whether in the proof hinted at above, there was an assumption that E and D are tangent to the inscribed circle, as I don't believe they will always be. If that's not the assumption, I'd very much appreciate an additional hint. Thanks much. - Avid

Your way of speaking about tangency is slightly confused. You can't speak, as you do, of "E and D" being "tangent to the inscribed circle" -- because points can be tangent to a circle; only lines (or line segments or rays) can. The inscribed circle is, by definition, tangent to each of the sides of the triangle. All you need to use in the proof is the fact that CE and CD are tangent to the inscribed circle. And this does not need proof; it is part of the definition of the inscribed circle. (Of course what does make for a significant assumption is that the angle bisectors all intersect at the incenter of the inscribed circle. But what's why I listed it as an assumption.)

There's a typo in my comment: it should so "points

can'tbe tangent to a circle; only lines (or line segments or rays) can."Thanks for the correction. I should have said that it isn't clear to me that D and E are points on the inscribed circle, unless the triangle is equilateral.

Whoops, you're absolutely right! Very careless of me. Even trained math people make silly mistakes! The problem is actually much harder, and it was perfectly fair to wonder whether there is an elementary solution. It turns out there is -- although there are also more advanced solutions (and the more math you've had, the faster the proof can go). Here is a compilation of proofs:

http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/steiner-lehmus

The theorem is known as the Steiner-Lehmus theorem. Thanks for posting this interesting problem here!

Thanks, Matt, this is a perfect answer!