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.

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:

the answer is a

step-by-step explanation:

because your just changing the position of where the shape is

e) none of the above because the measure of angle cba is 101°