Question

Question 4 (ii)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that:
(ii) ar(APD) + ar(PBC) = ar(APB) +ar(PCD)

Answer 1

Let us draw a line segment MN, passing through point P and parallel to line segment AD. In parallelogram ABCD,
MN|| AD (By construction) ... (6)
ABCD is a parallelogram.
$\therefore$ AB || DC(Opposite sides of a parallelogram)
$\Rightarrow$ AM || DN... (7)
From equations(6) and(7), we obtain,
MN|| AD and AM || DN
Therefore, quadrilateral AMND is a parallelogram.
It can be observed that $\Delta APD$ and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.
$\therefore Area\left(\Delta APD\right)=\frac{1}{2}Area\left(AMND\right)...\left(8\right)$
Similarly, for $\Delta PCB$ and parallelogram MNCB,
$Area\left(\Delta PCB\right)=\frac{1}{2}Area\left(MNCB\right)...\left(9\right)$
Adding equations (8) and (9), we obtain,
$Area\left(\Delta APD\right)+Area\left(\Delta PCB\right)=\frac{1}{2}\left[Area\right(AMND)+Area(MNCB\left)\right]$
$Area\left(\Delta APD\right)+Area\left(\Delta PCB\right)=\frac{1}{2}Area\left(ABCD\right)...\left(10\right)$
On comparing equations (5) in the previous question and (10), we obtain,
$Area\left(\Delta APD\right)+Area\left(\Delta PBC\right)=Area\left(\Delta APB\right)+Area\left(\Delta PCD\right)$