<Tri-2024-2097>, hint

Suppose line EP cuts C in Q and segment CPF cuts AD in R

Apply  Menelaus theorem to triangle CEQ with traverse BPG and triangle  BAD with

traverse CPF cuts D in R

CB/CE xEP/PG x QG/GC=1, CB/BD xDR/RA x AF/FB=1

Let QG/GC=y/(1-y), AF/FB=x/(1-x) Since EPQ and ARD are parallel

EP/PG=DR/RA, and CE=BD.   Hence QG/GC=AF/FB  y/(1-y)=x/(1-x). x=y, 1-x=1-y GC=FB.