Question

In a parallelogram A B C D, E and F are points on DC and AD respectively. Proof that area of triangle BFC is equal to area of triangle AEB ​

Answer 1

Answer:

Answer:

Given: ABCD is a parallelogram and E & F are points on side CD & AD

To prove: ar ( ΔAEB ) = ar ( Δ BFC )

Figure is attached

Proof,

ABCD is parallelogram means AB is parallel to CD and AD is parallel to CB.

Δ AEB and Parallelogram ABCD are on Same Base AB and Between Same Parallel Lines .i.e., AB & CD

Now According to a theorem which states that If a triangle and a parallelogram are on same base and between same parallels, then area of triangle is equal to half of area of parallelogram

⇒ ar (ΔAEB) = \frac{1}{2}\:ar (ABCD)

2

1

ar(ABCD) ................. (1)

Δ BFC and Parallelogram ABCD are on Same Base BC and Between Same Parallel Lines .i.e., BC & AD

Now, by same theorem,

⇒ ar (ΔBFC) = \frac{1}{2} \:ar (ABCD)

2

1

ar(ABCD) ................. (2)

From equation (1) & (2), we get

ar ( Δ AEB ) = ar ( Δ BFC )

Hence Proved