Question

$P$ is a point in the interior of a parallelogram $ABCD$. Show that $ar\left(△APB\right)+ar\left(△PCD\right)=\frac{1}{2}ar\left(ABCD\right)$

Answer 1

ABCD is a $\parallel$gm, in which AB$\parallel$CD & AD$\parallel$BC. Draw a line EF$\parallel$AB$\parallel$CD passing through P. such that AE$\parallel$BF & AB$\parallel$EF. Hence, EFBA is $\parallel$gm. Similarly EFOD is also a $\parallel$gm. Now, $△$APB and $\parallel$al lines AB and EF, therefore

ar $△$APE = $\frac{1}{2}arEFBA⟶\left(1\right)$

Similary ar $△$PCD = $\frac{1}{2}arEFCD⟶\left(2\right)$

Adding equation (1) and (2), we get

ar $△$APB + ar $△$PCD = $\frac{1}{2}$(ar EFBA + ar EFCD)

ar $△$APB + ar $△$PCD = $\frac{1}{2}$ ar $\parallel$gm ABCD