Consider a quadrilateral \(ABCD\) . Find the necessary and sufficient condition with proof so that there exists a point \(P\) in the interior of \(ABCD\) such that \(A(PAB)=A(PBC)= A(PCD)= A(PDA)\).

\(\text{A( ) represents Area}\)

Nice solutions are always welcome!

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## Comments

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TopNewestSuch a point \(P\) exists if and only if one diagonal bisects the area of the quadrilateral.

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Can you please provide the proof sir ?

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Sir, thanks for your answer.But it may be more beneficial for us if you post a nice proof.

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Now that you know the right condition, you should try proving it.

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Sir, actually this problem is from the homework of our class, and our sir has a different condition- P exists iff one diagonal bisects the other.

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Yeah , thats why I posted this as a note.

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In a quadrilateral, the conditions "one diagonal bisects the area" and "one diagonal bisects the other diagonal" are equivalent.

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@Calvin Lin @Brian Charlesworth @Trevor Arashiro@Pi Han Goh @Chew-Seong Cheong @Mehul Arora @Sharky Kesa @Sanjeet Raria @Sudeep Salgia @Ronak Agarwal @everyone help us.

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Nihar, Tujhe pata hai bhai ki Mujhe Geometry nahi aati :3 Jo Bhi Ho. Thanks For @Mentioning Me :D

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I mentioned you so that you will have something interesting in geometry. ;)

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me , @Kalash Verma @Harsh Shrivastava @CH Nikhil are in need of a nice solution.Thanks!

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