It's not necessary to deal with the neighbors at all. If T is where the bisector of angle QYZ intersects the bisector of QXR, then it will also be the point where the bisector of YZR will intersect the bisector of QXR. Here is why:

(i) The angle bisector of QYZ is the set of all points that are equidistant from YQ and YZ.

(ii) The angle bisector of YZR is the set of all points that are equidistant from YZ and ZR. (A pair of perpendiculars to YZ and ZR are shown.)

(iii) Therefore, the point where the two bisectors intersect is a point that is equidistant from ZR and YQ.

(iv) Such a point must lie on the bisector of YXZ. So, the point of intersection of all three bisectors must agree at the maple tree M.

Map of Solution

This elegant proof is due to a 14-year-old boy named Tyler.

The essence of this problem comes from the nice elementary geometry text by Edwin Moise and Floyd Downs entitled Geometry (ISBN 0-201-25335-6, Addison-Wesley 1991).