In DABC, let CL be the angle bisector of angle C, M the midpoint of AB, and MD||CL. (D lies on the longest of AC and BC.) Then points M and D split the perimeter of DABC into equal halves.
Assume AC > BC. Extend AC beyond C to F such that, BC = CF and, consequently, AF = AC + BC. Further, DBCF is isosceles, its angle bisector at C is perpendicular to the base BF. On the other hand, it is also perpendicular to the external angle bisector, i.e. CL, from which CL||BF. Since M is the midpoint of AB and MD||CL, we also have MD||BF, which implies that D is the midpoint of AF in DAFB:
