<Tri-2024-2092?>. hint

1>  L,M,N are centroids of triangles PBC, PCA, PAB [PELF]=[PEL]+[PFL]=1/3 [PBC]

2.  Suppose PL,PM,PN cut sides BC, CA, AB, in midpoints of BC,CA,AB, say L',M',N' then triangles L'M'N' is analogous to ABC

and since PL=2/3 PL' PM=2/3 PM' PN=2/3 PN'. LMN is analogous to L'M'N' and ABC

P is the center of homothety of triangles  ABC and LMN. 
Hence DL, EM, FN are concurrent