Question

$P$ is a point in the interior of a parallelogram $ABCD$. Show that $ar\left(△APD\right)+ar\left(△PBC\right)=ar\left(△APB\right)+ar\left(△PCD\right)$

Answer 1

ABCD is a $\parallel$gm such that AB$\parallel$CD & AD$\parallel$BC. Draw line GM passing through P, parallel to AD & BC. i.e. GH$\parallel$AD$\parallel$BC.

Here, GH$\parallel$AD (By construction) & AG$\parallel$DH ($\because$ AB$\parallel$CD & G & H are points on AB & CD respectively).

$\therefore$ AGHD is a $\parallel$gm.

Now, $△$APD and $\parallel$gm AGHD are on the same base AD and between the bank $\parallel$als AD & Gm.

Therefore,

ar $△$APD = $\frac{1}{2}$ar (AGHD) $⟶\left(1\right)$

Similarly ar ($△$PCB) = $\frac{1}{2}$ar (GBCH) $⟶\left(2\right)$

Adding equations (1) & (2), we get

ar $△$APD + ar $△$PCB = $\frac{1}{2}$ (ar AGHD + ar GBCH)

$\Rightarrow$ ar $△$APD + ar $△$PCB = $\frac{1}{2}$ ar ABCD

Also, $\frac{1}{2}$ ar ABCD = ar $△$APB + ar $△$PCD

$\therefore$ ar $△$APD + ar $△$PBC = ar $△$APB + ar $△$PCD