Question

a point T is taken on the side PQ of the parallelogram PQRS, and line ST and RQ are produced to meet at V. prove that the triangles VSQ and VTR are equal in area.​

Answer 1

Step-by-step explanation:

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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