Question

A point $ E$ is taken on the side $ BC$ of a parallelogram $ ABCD. AE$ and $ DC$ are produced to meet at $ F.$ Prove that ar $ (△ADF)=$ ar $ \left(ABFC\right)$

Answer 1

Given: $ABCD$ is a parallelogram.

Since,Triangles on the same base and between the same parallels are equal in area.

So,$△ABC$ and $△ABF$ are on the same base $AB$ and between the same parallels, $AB\parallel CF$

$\Rightarrow$area $(△ABC)=$ area $(△ABF)eq\left(1\right)$

We also know that, The diagonal of a parallelogram divides it into two triangles of equal area

$\Rightarrow$area $(△ABC)=$ area$(△ACD)eq\left(2\right)$

Now, area $(△ADF)=$ area$(△ACD)$ $+$ area $\left(ACF\right)$

$\therefore$ area $(△ADF)=$ area $(△ABC)+$ area $(△ACF)$ $($ From $eq\left(2\right))$

$\Rightarrow$ area $(△ADF)\hspace{0.17em}=$ area $(△ABF)+$ area$(△ACF)$ $($From $eq\left(1\right))$

$\Rightarrow$ area $(△ADF)=$ area $\left(ABFC\right)$

Hence, area $(△ADF)=$ area $\left(ABFC\right)$ proved.