Question

In the following figure, $BD$ is parallel to $CA,E$ is midpoint of $CA$ and $BD=\frac{1}{2}CA.$ Prove that:
$Ar.\left(△ABC\right)=2\times Ar.\left(△DBC\right)$

Answer 1

It is known that if one pair of opposite sides of a quadrilateral are equal and parallel then the quadrilateral is a parallelogram.

Here $BD=CE$ and $BD\parallel CE.$ thus $BCED$ is a parallelogram.

$Ar.\left(△DBC\right)=Ar.\left(△EBC\right)$....(Since they have the same base and are between the same parallels)

In $△ABC,$
$BE$ is the median,

So, $Ar.\left(△EBC\right)=\frac{1}{2}Ar.\left(△ABC\right)$

Hence, $Ar.\left(△ABC\right)=2Ar.\left(△EBC\right)$

$\Rightarrow Ar.\left(△ABC\right)=2Ar.\left(△DBC\right)$