Let #|ABC|# denote area of #DeltaABC#
Let #|ABC|=12a#
given #AE=EB, => |CAE|=|CEB|=(12a)/2=6a#
as #BD:DC=2:1, => |DEB|:|CEB|=2:1#,
#=> |DEB|=4a, => |CED|=2a#
let #F and G# be the midpoint of #AC and AD#, respectively,
#=> EGF# // #BC, => EG:GF=BD:DC=2:1#
let #BD=2x, => DC=x#
# => EG=x, GF=1/2x#
as #EG=DC=x, and EG# // #DC => GC# // #ED#,
#=> EDCG# is a parallelogram of which #GD and EC# are the diagonals, #=> |EDCG|=2*|CED|=2*2a=4a#
recall that the diagonals of a parallelogram divides it into 4 equal areas,
#=> |PED|=|EDCG|/4=(4a)/4=a#
#=> |PED|/|ABC|=a/(12a)=1/12#
