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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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The angle bisectors aren't typically perpendicular to the sides. In particular the incenter I will not be on any of the perpendiculars you mention. –  coffeemath Oct 18 '12 at 3:58
    
I think you want to start by defining a point J to be where two of the bisectors meet. That way you aren't using circular logic. Then using only that point J and what you have so far, you try to draw other things like subtriangles. Hunt around for maybe some similar or congruent shapes, and hopefully prove that the line from the third vertex through J does in fact bisect the angle there. This approach will not be circular, and after done, only then call the J by the name incenter, and change the J to an I. –  coffeemath Oct 18 '12 at 4:01
    
You probably mean that point $I$ is on the bisector of $\angle A$. Then what you have in the box is fine, but this is correct for any point along the bisector, so $I$ need not be on the bisector of one of the other angles. DonAntonio's argument deals with this. –  Ross Millikan Oct 18 '12 at 4:01
    
I think you want to start by defining a point J to be where two of the bisectors meet. That way you aren't using circular logic. This is okay? Assuming that two bisectors meet is okay right? I'll fix up the start up then –  jip Oct 18 '12 at 4:04
    
@coffeemath, I am making perpendiculars after I connect the point from the bisectors. –  jip Oct 18 '12 at 4:10

1 Answer 1

up vote 3 down vote accepted

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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Could you look at my newly edit proof? –  jip Oct 18 '12 at 4:18
    
I think it's fine but it is basically the same as above and perhaps with too many words: any point on an angle bisector is at the same distawnce from both legs...period. No need, imo, to mention all that perpendiculars stuff, which is obvious and makes the proof a little cumbersome. –  DonAntonio Oct 18 '12 at 4:24
    
My professor actually told us to do that. How would you condense the proof (no pictures please)? –  jip Oct 18 '12 at 4:34
    
Just as I wrote it, making sure it is well understood what the definition of angle bisector as a locus of points is. –  DonAntonio Oct 18 '12 at 4:47
    
But aren't we trying to prove the definition? It seems like you just stated what it is. –  jip Oct 21 '12 at 5:33

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