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 $\blacksquare$