In the adjoining figure, $B D \| C A, E$ is the midpoint of $C A$ and $B D=\frac{1}{2} C A$. Prove that $\operatorname{ar}(\triangle A B C)=2 \operatorname{ar}(\Delta D B C)$.
E is the midpoint of CA.
So, AE = EC .....(1)
Also, $\mathrm{BD}=\frac{1}{2} \mathrm{CA}$ (Given)
So, BD = AE .....(2)
From (1) and (2) we have
BD = EC
BD || CA and BD = EC so, BDEC is a parallelogram
$B E$ acts as the median of $\triangle A B C$
so, $\operatorname{ar}(\triangle \mathrm{BCE})=\operatorname{ar}(\triangle \mathrm{ABE})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$ .......(i)
$\operatorname{ar}(\triangle D B C)=\operatorname{ar}(\triangle B C E)$ .....(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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