Question

E and F are two points on the sides DC and AD of the parallelogram ABCD.prove that area of triangle AEB=area of tringle BFC.

Answer 1

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) = $\frac{1}{2}\:ar (ABCD)$ ................. (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)

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

ar ( Δ AEB ) = ar ( Δ BFC )

Hence Proved

Answer 2

Answer:

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Step-by-step explanation:

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