Question

$A$ point $E$ is taken on the side $BC$ of a parallelogram $ABCD,AE$ and $DC$ produced to meet at $F$. Prove that area $\left(\Delta ADF\right)=$ Area $\left(l{l}^{m}ABFC\right)$.

Answer 1

Given: $ABCD$ is a parallelogram. $A$ point $E$ is taken on the side $BC,AE$ and $DC$ are produced to meet at $F$.

Proof : Since $ABCD$ is a parallelogram and diagonal $AC$ divides it into two triangles of equal area, we have

Refer image,

$ar\left(\Delta ADC\right)=ar\left(\Delta ABC\right)$..........(1)

As $DC||AB$, So $CF||AB$

Since triangles on the same base and between the same parallels are equal in area, so we have

$ar\left(\Delta ACF\right)=ar\left(\Delta BCF\right)$..........(2)

Adding (1) and (2), we get

$ar\left(\Delta ADC\right)+ar\left(\Delta ACF\right)=$ $ar\left(\Delta ABC\right)+ar\left(\Delta BCF\right)$

$\Rightarrow ar\left(\Delta ADF\right)=ar\left(ABFC\right)$

Hence proved