Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB)×ar(CPD)=ar(APD)×ar(BPC)

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

1.	Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB)×ar(CPD)=ar(APD)×ar(BPC)
Answer» Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that $a r (A P B) \times a r (C P D) = a r (A P D) \times a r (B P C)$