Question

In the adjoining figure, $BD||CA,E$ is the midpoint of $CA$ and $BD=\frac{1}{2}CA$. Prove that $ar\left(△ABC\right)=2ar\left(△DBC\right)$.

Answer 1

It is given that $E$ is the midpoint of $CA$ and $BD=\frac{1}{2}CA$

So we get

$BD=CE$

We know that $BD\parallel CA$

$BD=CE$

In the same way $BD\parallel CE$

$BD=CE$

Hence, $BCED$ is a parallelogram.

We know that $△DBC$ and $△EBC$ lie on the same base and between the same parallel lines

So we get

Area of $△DBC$= Area of $△EBC$.....(1)

From the figure we know that $BE$ is the median of $△ABC$

We get

Area of $△BEC=\frac{1}{2}$ ( Area of $△ABC$)

Using equation (1) we get

Area of $△DBC=\frac{1}{2}$ ( Area of $△ABC$)

So we get

Area of $△ABC=2$ ( Area of $△DBC$)

Therefore it is proved that $ar\left(△ABC\right)=2ar\left(△DBC\right)$.