Point e is the midpoint of side bc of parallelogram abcd (labeled counterclockwise) and ae ∩ bd =f. find the area of abcd if the area of △bef is 3 cm2.

Explanation:

Since here ABCD is the parallelogram.

Where E is the mid point of the line segment BC.

And, F is the intersection point of the segments AE and BD,

Also,  Area of △BEF is 3 .

We have to find Area of parallelogram ABCD.

Since,In  ΔBEF  and ΔACD,

∠ADF=∠EBF ( because AD ║ BE )

∠DFA = ∠BFE (vertically opposite angle)

Thus, By AA similarity postulate,

,

Thus,

But, AD = 2 BE,

Therefore,   = 4/1

Thus, area of Δ ADF =

Similarly,

Now, let the area of the triangle AFB = x

Thus, x + 12 = 2(x+3) ( because the area of the ΔADB = 2(area of ΔBAE) )

⇒ x = 6

Therefore, area of ΔAFB= 6

⇒ area of ΔABD = 12 + 6 = 18

By the definition of diagonal of parallelogram,

Area of parallelogram ABCD = 2× area of Δ ABD= 2 × 18 = 36