IF P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.

In || gm ABCD, P is any point in the || gm
AP and BP are joined
To prove : ar $\left(△APB\right)<\frac{1}{2}ar\left(||gmABCD\right)$
Construction : Draw $DN\perp ABandPM\perp AM$

Proof : $ar\left(||gmABCD\right)=base\times altitude=AB\times DN$
From (i) and (ii)
DN > PM or PM < DN
$\Rightarrow AB\times PM<AB\times DN\Rightarrow AB\times PM<\frac{1}{2}AB\times DN\Rightarrow ar\left(△PAB\right)<\frac{1}{2}ar||gm\left(ABCD\right)$