Converse Proof

[PiRsq]

Prove that the bisectors of the angles of any triangle ABC are concurrent. The point of intersection is called the incentre, I.

Proof:

-Angle A and B are bisected and their bisectors meet at a point I
-Assume a line segment bisects angle C and meets the bisector of angle A at point O
-Assume a line from O to B that will bisect angle B
-But angle B is already bisected by segment BI
-So line segment OB must be BI
-Thus point O is I
-Therefore the incentre is at I


Does this proof workout?

[arcnets]

I think it does not work.
Look:

-Angle A and B are bisected and their bisectors meet at a point I
-Assume a line segment bisects angle C and meets the bisector of angle A at point O
OK, so far.
-Assume a line from O to B that will bisect angle B

This is a false assumption:
You have already defined I to be on the bisection of angle B.
So you may not assume that O, which has already been defined otherwise, is also on the bisection of angle B.

Since the assumption is false on the basis of your proposition, it does not lead to the conclusion that the proposition is false.

[arcnets]

Here's a proof:

http://aleph0.clarku.edu/~djoyce/java/elements/bookIV/propIV4.html