Diagonals AC and BD of a quadrilateral ABCD 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

Let us draw $A M ⊥ B D$ and $C N ⊥ B D$

WE know that, Area of a traingle $= \frac{1}{2} \times B a s e \times A l t i t u d e$
$a r (A P B) \times a r (C P D)$
$= [\frac{1}{2} \times B P \times A M] \times [\frac{1}{2} \times P D \times C N]$
$= \frac{1}{4} \times B P \times A M \times P D \times C N$ --------(i)

$a r (A P D) \times a r (B P C)$
$= [\frac{1}{2} \times P D \times A M] \times [\frac{1}{2} \times C N \times B P]$
$= \frac{1}{4} \times P D \times A M \times C N \times B P$
$= \frac{1}{4} \times B P \times A M \times P D \times C N$ -------(ii)

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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a r (A P B) \times a r (C P D) = a r (A P D) \times a r (B P C)

a r (A P B) \times a r (C P D) = a r (A P D) \times a r (B P C)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that

[Hint: From A and C, draw perpendiculars to BD]

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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