Given: Triangle ABC

Show: The angle bisectors of all three sides of triangle ABC pass through a single point

P.
 
 

Proof: (For clarity, this proof will make use of Lemma 2 below.)

Construct rays l, m, and n that are the angle bisectors of angles ABC, BCA, and BAC respectively. Since a side of each angle ABC and BCA contains the segment BC, their respective angle bisectors l and m are not parallel. So l and m intersect at some point P.

Since P lies on l, the angle bisector of angle ABC, P is equidistant from the sides BA and BC by Lemma 2. Since P lies on m, the angle bisector of BCA, P is also equidistant from the sides CB and CA by Lemma 2. So P is equidistant from the sides AB and AC.
 
 

Then P must also lie on n, the angle bisector of BAC, by Lemma 2. Since P lies on l, m, and n, these three angle bisectors are concurrent.

Lemma 2

The angle bisector of an angle XYZ is the set of all points that have equal (perpendicular) distances to rays YX and YZ.