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# Why are we caring so much about the existence of such a tiny Point P?

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!

Note by Nihar Mahajan
1 year, 11 months ago

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Such a point $$P$$ exists if and only if one diagonal bisects the area of the quadrilateral. · 1 year, 11 months ago

Can you please provide the proof sir ? · 1 year, 11 months ago

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. · 1 year, 10 months ago

In a quadrilateral, the conditions "one diagonal bisects the area" and "one diagonal bisects the other diagonal" are equivalent. · 1 year, 10 months ago

Oh yes, thank you for pointing that out to me · 1 year, 10 months ago

Yeah , thats why I posted this as a note. · 1 year, 10 months ago

Did you get a proof · 1 year, 10 months ago

I got the proof for the condition- P exists iff one diagonal bisects the other.I did not get proof for the condition that Jon Hausmann has posted. · 1 year, 10 months ago

yes,same · 1 year, 10 months ago

Sir, thanks for your answer.But it may be more beneficial for us if you post a nice proof. · 1 year, 11 months ago

Now that you know the right condition, you should try proving it. · 1 year, 11 months ago

me , @Kalash Verma @Harsh Shrivastava @CH Nikhil are in need of a nice solution.Thanks! · 1 year, 11 months ago

Nihar, Tujhe pata hai bhai ki Mujhe Geometry nahi aati :3 Jo Bhi Ho. Thanks For @Mentioning Me :D · 1 year, 11 months ago

I mentioned you so that you will have something interesting in geometry. ;) · 1 year, 11 months ago