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

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

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Solution

IN quadrilateral ABCD, diagonal AC and BD intersect each other as P
To prove : ar $(△ A P B) \times a r (△ C P D) = a r (A P D) \times a r (△ B P C)$
Construction : Draw Al and CN perpendiculars on BD

Proof: $a r (△ A P D) \times a r (△ B P C) = (\frac{1}{2} \times A L \times D P) \times (\frac{1}{2} \times C N \times B P) = (\frac{1}{2} \times B P \times A L) \times (\frac{1}{2} \times D P \times C M) = a r (△ A P B) \times a r (△ C P D)$

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Similar questions

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)

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