<Tri-2024-2141>, hint

Let H be the intersection of AB and CF extended  the CF=FH and DA=DC

FE//AB. Let prove GE is parallel to BC .

Let B(0,0), C(3,0), A(3,4). Then CE=4x(3/(5+3))=3/2, BE: y=1/2 x

CG: y=-2(x-3), BD:=2/3 x. Hence G(9/4, 3/2) Hence GE is parallel to BC

<DFE=<ABE =1/2 <ABC, <GEF=<EBC=1/2 <ABC. Let Q be the intersection of DF wit EG

Triangle QFE is isosceles and <GFE=90  Triangle GFE is right triangle

 Hence GQ=QE.