Let x=∠JFH=∠HFG (FH bisects ∠JFG)
∠JEF=∠HFG=x (corresponding angles between parallels EJ and FH)
∠EJF=∠JFH=x (alternate angles between parallels EJ and FH)
ΔEFJ is isosceles because ∠JEF=∠EJF, so EF=JF, therefore 0.8y+12=2y-12, 24=1.2y, so y=20, EF=JF=40-12=16+12=28. EJ=y+6=26. Perimeter=28+28+26=82 units.