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.To Prove : Construction :From A to C , DRAW perpendiculars AE and CF respectively to BD.Proof : : : : :From (1) and (2),__________________________

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

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