The applet may suggest the following statment:

Assume AC > BC. Extend AC beyond C to F such that, BC = CF and, consequently, AF = AC + BC. Further, ΔBCF 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 ΔAFB:

By construction,

so that