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I am trying to prove that the internal bisectors of the angles of a triangle meet at a point - the incenter.

I need someone to critique this incomplete proof for me

Consider $\triangle ABC$ with angle bisectors $\angle A, \angle B, \angle C$. Assume a point I is on $\angle A$ and inside the triangle. Drop perpendiculars from I to side AC and AB to points Z and Y respectively. Thus we have $\triangle YAI \cong \triangle ZAI$ (AAS) $\implies$ IZ = IY

So my idea is to extend this idea to the other vertices and make IX = IY and IZ and finish off the proof. But I am not sure if this is considered circular logic. One of my friend started off like me and after he wrote down IZ = IY, he begins to say something along the lines of "Since I lies on the angle bisector of B..." and he pretty much repeated the same procedure and finished his proof.

But is it okay to "assume" that the same point I is also lying on some other bisector? I thought about using another "point" like I' and somehow show that I' = I later on. BUt that seems too difficult.

The picture is just an idea. I won't include it in my proof (I think my start up gives the reader a good idea of the triangle construction)

EDIT: Refined Proof

Consider $\triangle ABC$ with angle bisectors $\angle A, \angle B, \angle C$. Assume a point I lies on two angle bisectors, say $\angle A$ and $\angle B$, and inside the triangle. Drop perpendiculars from point I to side AC and AB to points Z and Y respectively. Thus we have $\triangle YAI \cong \triangle ZAI$ (AAS) $\implies$ IZ = IY. Similiarily, drop perpendiculars from point I to the point X on side CB and we obtain $\triangle XCI \cong \triangle ZCI$ (AAS) $\implies IZ = IX$. By transitivity, we have $IY = IZ = IX$. Hence point I is equidistant from all three sides of the triangle and is the incenter $\blacksquare$

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    Yes, didn't notice that. It then is fine as long as$I$is not assumed to be on all three bisectors, as others have remarked.2012-10-18

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Of course it is not correct to assume I lying in two different bisectors: this is part of what must be proved!

I'd go like this: let I be the intersection point of two angle bisectors, say $\angle A\,,\,\angle B\,$ .

We get at once that the point I is at the same distance of the three sides of the triangle and, thus, it is ALSO on the third angle's bisector!

Remember: the bisector of an angle is the locus of all point that are at the same distance from both angle's legs.

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    No, you are trying to prove that all three bisectors of a triangle's angles meet in a single point. For that, according to my solution, one already needs to know what the definition of "angle bisector" is a locus of points.2012-10-21