Question

ABCD is a parallelogram and E is any point on AB . If DE produced meets CB produced at F , prove that-----

i) the area of triangle ADF = area of the triangle DEC

ii) the area of triangle AEF = area of triangle BEC.

Answer 1

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= $\frac{1}{2}$ × (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)=  $\frac{1}{2}$ × 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)=  $\frac{1}{2}$ × 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)=  $\frac{1}{2}$ × 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)

Answer 2

Step-by-step explanation:

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= \frac{1}{2}

× (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)= \frac{1}{2}

2

1

× 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)= \frac{1}{2}

2

1

× 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)= \frac{1}{2}

2

1

× 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)