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.

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Solution

In || gm ABCD, P is any point in the || gm
AP and BP are joined
To prove : ar $(△ A P B) < \frac{1}{2} a r (| | g m A B C D)$
Construction : Draw $D N ⊥ A B a n d P M ⊥ A M$

Proof : $a r (| | g m A B C D) = b a s e \times a l t i t u d e = A B \times D N$
From (i) and (ii)
DN > PM or PM < DN
$\Rightarrow A B \times P M < A B \times D N \Rightarrow A B \times P M < \frac{1}{2} A B \times D N \Rightarrow a r (△ P A B) < \frac{1}{2} a r | | g m (A B C D)$

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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.

△ A P B + △ C P D

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a r (△ A P B) + a r (△ P C D) = \frac{1}{2} a r (A B C D)

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