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A point D is taken on the side BC of a D...

A point D is taken on the side BC of a `DeltaABC` such that `BD = 2DC.` Prove that `ar(DeltaABD) = 2 ar (DeltaADC)`

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In `DeltaABC`, we have `BD = 2DC`
Let E is the mid-point of BD
Then, `BE = ED = DC`
Since, AE is the median of `DeltaABD`
`:. Ar(DeltaABE) = ar(DeltaAED)`
(median divides the traingle into two equal areas) ...(1)
Also, since AD is the median of `Delta AEC`,
`:. ar (DeltaAED) = ar(DeltaADC)`
(median divides the triangle into two equal areas)...(2)
From eqs. (1) and (2), we have
`ar(DeltaABE) = ar(DeltaAED) = ar(DeltaADC) = x`(say)
Now, `ar(DeltaABD) = ar(DeltaABE) + ar(DeltaAED) = x + x = 2x = 2 xx ar (DeltaADC)`
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