(N/A) Let the quadrilateral plot be $ABCD$. Let $E$ be the point where the diagonal $AC$ intersects the boundary of the portion to be taken. To implement the proposal,we draw a line through $D$ parallel to $AC$,which meets the extended side $BC$ at point $F$.
Now,consider $\Delta DAF$ and $\Delta DCF$. These triangles are on the same base $DF$ and between the same parallel lines $AC$ and $DF$.
Therefore,$\text{ar}(\Delta DAF) = \text{ar}(\Delta DCF)$.
Subtracting $\text{ar}(\Delta DEF)$ from both sides,we get:
$\text{ar}(\Delta DAF) - \text{ar}(\Delta DEF) = \text{ar}(\Delta DCF) - \text{ar}(\Delta DEF)$
$\Rightarrow \text{ar}(\Delta ADE) = \text{ar}(\Delta CEF)$.
This means the area of the triangular portion $\Delta ADE$ (which the Panchayat takes) is equal to the area of the triangular portion $\Delta CEF$ (which is given to Itwaari). By adding $\Delta CEF$ to the remaining part of the plot,Itwaari gets a new triangular plot $\Delta ABF$.
To verify the area: $\text{ar}(\Delta ABF) = \text{ar}(ABCE) + \text{ar}(\Delta CEF)$.
Since $\text{ar}(\Delta CEF) = \text{ar}(\Delta ADE)$,we have:
$\text{ar}(\Delta ABF) = \text{ar}(ABCE) + \text{ar}(\Delta ADE) = \text{ar}(\text{quadrilateral } ABCD)$.
Thus,the total area remains the same.