<Polygon-2024-265>, hint

Since areas of ABC and BCD are equal AD and BC are parallel 

Similarly  CD and BE are parallel. Let F be the intersection , then

ABCF is parallelogram , whose area is 2[ABC], [CDE]=[ABC]

[AEF]+[DEF]=[ADE]=[ABC]. Hence 3[ABC]<ABCDE]<4[ABC]