<Tri-204-2144>, hint

suppose line MN cuts line BC in K and  <LMK'=90 K' on line BC.

By Menelaus to triangle ABC  with traverse MNK 

(AM/MN)(AN/NB)(BK/KC)=1. (AM/MA)(AN/NB)(BL/LC)=1 is given

Hence BK/KC=BL/LC ---> (B,C: L,K). Since AL, AK' are internal, external bisectors of 

<CAB (B,C :L,K')---> K=K' Hence <LMK =<LMN=90

ML, MK are internal and external bisectors of <CMB.