<Tri-224-2099>, hint

Since triangles AEF and ANM are similar EF//NM.

 Suppose  line parallel to NM, through D, cuts AB in G and prove

AG+AF=2AM. AF=b cosA, AM= b sinB sinC,  AG/ sin ADG=AD/sin AGD

<ADG=90-C+B. AG=b cos(C-B).

Hence AG+AF=b cos(C-B)+b cosA=2b siBsinC=2AM