Geometry problem??

Question

For triangle ABC, let AD and BE be the bisectors of angles CAB and CBA.
AD = BE. Prove that ABC is isoceles.

It's the LENGTHS AD and BE that are the same, not angles. AD bisects angle CAB, and BE bisects angle CBA. That is not the same as saying angles CAB = CBA, it's something to be proven.

All right, so the proof of this could be horribly messy. Any brief proofs? A sketch of the proof will do, if that's all that can fit in here.

Answer 1

OK, I'll just post the link then.

http://www.mathpages.com/home/kmath433.htm

Answer 2

This is a problem that requires an indirect proof also called reduction to absurdity. Euclid used these proofs occaisionally.

Basically say that by the law of Trichotomy, either AC = BC, AC > BC or AC < BC.

Assume that AC > BC and show that this results in an absurdity, e.g AD not = DE. Then assume AD < BC and show this results in the same absurdity.

Therefore AC = BC as it is the only remaining possibility.

I leave the details to you. This is a very old problem and probably can be found on the internet somewhereif you get stuck. But I don't want to spoil your fun by giving you the detailed solution.

Happy solving.

Answer 3

If AD=BE, then angles CAB and CBA are also equal to each other. As the two angles are equal, the sides opposite them must also be equal, and this proves the triangle to be either isosceles or equilateral.

Answer 4

bisector means it cuts the angle in half.

if bisector are equal, then angles CAB and CBA are equal. if two angles are equal then ABC is isosceles.