![enter image source here]()
Let |ABC||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
