<Tri-2025-2208>, hint

Suppose P on a median AA'. Then by Ceva's AC'/C'B x BA'/A'C x CB'/B'A=1

Since BA'=A'C,   AC'/C'B=AB'/B'C , B'C'//BC Hence [PBA']=[PCA']=p, [B'BC]=[C'CB]-->

[PB'C]=[PC'B]=q and so [PAB']=[PAC']=r Hence p+q+r=1/2 [ABC]