In the given figure, ABCD is a parallelogram. E is any point on AB. CB and DE are produced to meet at F. Prove that area of triangle CEF = area of triangle BDF. if AB=EB, then area triangle BCE = area of triangle BDE = 1/4 area of parallelogram ABCD.

Answer:Keep these axioms in mind while proving this 1. If a triangle and parallelogram have same base and are between the same parallel, then area of Parallelogram=  × (Area of triangle) 2. Two parallelogram sharing the same base and if the sides parallel to the base are lying in the same line, their areas are equal. Given: A B CD is a parallelogram and E is any point on AB . If DE produced meets CB produced at F . Proof: AS, A B CD is a parallelogram. AB ║CD AD ║CB C F ║AD Δ A D F and ║gm ABCD have same base AD and lying between the same parallels, AD ║CF. →→Area ( Δ A D F)=   × Area (║gm ABCD)------(1) Similarly, Δ  D E C and ║gm ABCD have same base CD and lying between the same parallels, CD ║AB. →→Area ( Δ  D E C)=   × Area (║gm ABCD)-----(2) From (1) and (2) →→→→Area ( Δ A D F)=Area ( Δ DEC) (2) Area ( Δ A D F)=  Area ( Δ DEC) →→Area ( Δ A D F)- Area (ΔA DE)=  Area ( Δ DEC)- Area (ΔA DE) →→Area ( Δ A E F)=Area ( Δ A D E)+Area ( Δ B E C)-Area (ΔA DE)→→[Area ( Δ  D E C)=   × Area (║gm ABCD)]→→→[ Area ( Δ  D E C)=Area ( Δ A D E)+Area ( Δ B E C)] →→Area ( Δ A E F)=Area ( Δ B E C) Step-by-step explanation: