If `P` is any point in the interior of a parallelogram `A B C D ,` then prove that area of the triangle `A P B` is less than half the area of parallelogram.

Explanation

To prove that area of triangle `APB` is less than half the area of the parallelogram.
Construction :Draw a parallelogram `ABCD`.
Draw `DN _|_AB` and `PM _|_AB`.
Now,

Area `( ABCD) = AB×DN`, ar `(/_\APB) =1/2(AB × PM)`

Now, ` PM`AB* PM `< `AB* DN`

⇒ `1/2(AB × PM) < 1/2 (AB×DN)`

⇒ `area (/_\APB)<1/2 area (Parallelogram ABCD)`.