Here we are with our good old friend, triangle ABC. The lines in that ugly shade of purple shows the angle bisectors of the angles at each vertice. D is the intersection of these lines, and the center of our focus as well. Grab A and try to position it in a way such that D lies outside of the triangle. Having a hard time? Try the same things with B. Still no luck? I hope that by now you can guess what will happen if we try to move C. Regardless of the position of these points or form of the triangle, the intersection of the angle bisectors will always be contained by the boundaries of ABC. Check out the proof below.

Theorem: The intersection of the angle bisectors of any triangle ABC will be contained within the boundaries of the triangle.
Proof: BF, CE, and AG are all angle bisectors. Show that D is an element of the half-plane containing A, the half-plane containing B, and the half-plane containing C.
1) D is an element of the half-plane containing A and with boundary BC
Note a: L and D lie on opposite sides of BC.
Note b: K and D lie on opposite sides of BC.
Because of a and b, we can assume that L and K are on the same half-plane, which we can see in the diagram as being outside the triangle. So D is in the half-plane using BC as a boundary and containing A.
2) D is an element of the half-plane containing B and with boundary AC
Same logic as above shows this is true.
3) D is an element of the half-plane containing C and with boundary AB
Same logic shows this is true.
Thus, D lies within the boundaries of ABC.