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A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F.

Prove that ar( ADF ) = ar (ABFC ).

Given : ABCD is a parallelogram. E is a point on BC. AE and DC are produced to meet at F.

To prove: area (ΔADF) = area (ABFC).

Proof :

area (ΔABC) = area (ΔABF)  ...(1) (Triangles on the same base AB and between same parallels, AB || CF are equal in area)

area (ΔABC) = area (ΔACD)    ...(2)  (Diagonal of a Parallelogram divides it into two triangles of equal area)

Now,

area (ΔADF) = area (ΔACD) + area (ΔACF)

∴ area (ΔADF) = area (ΔABC) + area (ΔACF)  (From (2))

⇒ area (ΔADF) = area (ΔABF) + area (ΔACF)  (From (1))

⇒ area (ΔADF) = area (ΔABFC)

 

:) :)

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Given : ABCD is a parallelogram. E is a point on BC. AE and DC are produced to meet at F. To prove: area (ΔADF) = area (ABFC). Proof : area (ΔABC) = area (ΔABF) ...(1) (Triangles on the same base AB and between same parallels, AB || CF are equal in area) area (ΔABC) = area (ΔACD) ...(2) (Diagonal of a Parallelogram divides it into two triangles of equal area) Now, area (ΔADF) = area (ΔACD) + area (ΔACF) ∴ area (ΔADF) = area (ΔABC) + area (ΔACF) (From (2)) ⇒ area (ΔADF) = area (ΔABF) + area (ΔACF) (From (1)) ⇒ area (ΔADF) = area (ΔABFC)
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