Question

Question 1
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that $ar\left(\Delta ADF\right)=ar\left(ABFC\right)$.

Answer 1

Construction: Join AC, BF and DE.
Proof:
AC is a diagonal of parallelogram ABCD.
$So,ar\left(\Delta ABC\right)=ar\left(\Delta ACD\right)$ ....(i)
Since, $\Delta$ ABF and $\Delta$ABC we are on the same base AB and between the same parallels AB and DF.
[since, AB || DC and DC produced to F]
$\therefore ar\left(\Delta ABF\right)=ar\left(\Delta ABC\right)$ .....(ii)
From eqs. (i) and (ii),
$ar\left(\Delta ABC\right)=ar\left(\Delta ACD\right)=ar\left(\Delta ABF\right)$ ...(iii)
On subtracting $ar\left(\Delta ABE\right)$ from both sides of eq. (ii), we get,
$ar\left(\Delta ABF\right)-ar\left(\Delta ABE\right)=ar\left(ABC\right)-ar\left(\Delta ABE\right)$
$\Rightarrow ar\left(\Delta BEF\right)=ar\left(\Delta AEC\right)$ ...(iv)
Now, $ar\left(AECD\right)=ar\left(ACD\right)+ar\left(AEC\right)$
$=ar\left(\Delta ABC\right)+ar\left(\Delta BEF\right)$ [from eqs. (i) and (iv)]
On adding ar $\Delta CEF$ on both sides, we get ,
$ar\left(AECD\right)+ar\left(\Delta CEF\right)$
$=ar\left(\Delta ABC\right)+ar\left(\Delta BEF\right)+ar\left(\Delta CEF\right)$
$\Rightarrow ar\left(\Delta ADF\right)=ar\left(ABFC\right)$