Question:
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.
Solution:
Draw DN ⊥ AB and PM ⊥ AB
Now,
ar (∥gm ABCD) = AB × DN, ar(ΔAPB) = (1/2) (AB × PM)
Now, PM < DN
⇒ AB × PM < AB × DN
⇒ (1/2)(AB × PM) < (1/2)(AB × DN)
⇒ ar(ΔAPB) < 1/2 ar(∥ gm ABCD)