Diagonals $A C$ and $B D$ of a quadrilateral $A B C D$ intersect each other at $P$ show that
$a r . (△ A P B) \times a r . (△ C P D) = a r . (△ A P D) \times a r . (△ B P C)$

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Solution

Data: Diagonals $A C$ and $B D$ of a quadrilateral $A B C D$ intersect each other at $P$ .
To Prove: $a r . (△ A P B) \times a r . (△ C P D) = a r . (△ A P D) \times a r . (△ B P C)$
Construction: Draw $A M ⊥ D B, C N ⊥ D B$
Proof: $a r . (△ A P B) \times a r . (△ C P D) =$
$= (\frac{1}{2} \times P B \times A M) \times (\frac{1}{2} \times P D \times C N)$
$= \frac{1}{4} \times P B \times A M \times P D \times C N . . . . (i)$
$a r . (△ A P D) \times a r . (△ B P C) =$
$= (\frac{1}{2} \times P D \times A M) \times (\frac{1}{2} \times P B \times C N)$
$= \frac{1}{4} \times P D \times A M \times P B \times C N . . . . (i i)$
From (i) and (ii)
$a r . (△ A P B) \times a r . (△ C P D) = a r . (△ A P D) \times a r . (△ B P C)$

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