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Question

In the adjoining figure, BD || CA, E is the midpoint of CA and BD = 12 CA. Prove that ar(∆ABC) = 2ar(∆DBC).

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Solution

E is the midpoint of CA.
So, AE = EC .....(1)
Also, BD = 12 CA (Given)
So, BD = AE .....(2)
From (1) and (2) we have
BD = EC
BD || CA and BD = EC so, BDEC is a parallelogram
BE acts as the median of ABC
so, ar(BCE) = ar(ABE) = 12arABC .....(1)
ar(DBC) = ar(BCE) .....(2) (Triangles on the same base and between the same parallels are equal in area)
From (1) and (2)
ar(∆ABC) = 2ar(∆DBC)

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