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Question

Question 6
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB)×ar(CPD)=ar(APD)×ar(BPC)  


Solution

Let us draw AMBD and CNBD.

We know that, Area of a triangle=12×Base×Altitude
ar(APB)×ar(CPD)
=[12×BP×AM]×[12×PD×CN]
=14×BP×AM×PD×CN --------(i)

ar(APD)×ar(BPC)
=[12×PD×AM]×[12×CN×BP]
=14×PD×AM×CN×BP
=14×BP×AM×PD×CN -------(ii)

From (i) and (ii),
ar(APB)×ar(CPD)=ar(APD)×ar(BPC)


 

Mathematics
NCERT Textbook
Standard IX

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