2000 ce maths mc Q54

point F is on AB, and point G is on CD , FG ⊥AB, FG ⊥ DC
area of △ABC = 1/2 (AB)(FG) = 4+5 =9, FG = 18/AB
area of △ABE = 1/2 (AB)(FE) =4, FE = 8/AB

∴ FG : FE = 18 :8 = 9:4
FE + EG : FE = 9:4
EG : FE = 5:4
∠ABE =∠CDE , ∠BAE =∠DCE (CORR. ∠s AB∥ DC)
∠AED = ∠CED (VERT. OPPO. ∠s)
∴ △ABE～△CDE (AAA)

∴area of △CDE/area of △ABE = [(EG)/(FE)]^2 = [5/4}^2 = 25/16
area of △CDE = 4(25)/16 = 6.25 CM^2

thank you very much!