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Diagnols of quad.ABCD meet at P. Prove that ar(APB)*ar(DPC)=ar(ADP)*ar(BPC)

Diagnols of quad.ABCD meet at P.


Prove that ar(APB)*ar(DPC)=ar(ADP)*ar(BPC)


 

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1 Answers

preeti prabhu
19 Points
9 years ago

Diagnols of quad.ABCD meet at P.

Prove that ar(APB)*ar(DPC)=ar(ADP)*ar(BPC)

solution:

construct altitude AM and CN such that M and N are points on diagonal BD.

Now using area of triangle = 1/2 * base *height,

ar(ABP) = AM*BP/2

ar(ADP) = AM*DP/2

ar(BCP)= BP*CN/2

ar(CPD) = DP*CN/2

ar(ABP *DCP) = AM*BP*DP*CN/4

ar(ADP* BPC) = AM*DP*BP*CN/4

from dis, we know, they are equal!

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