# 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
6 years, 2 months ago

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Such a point $P$ exists if and only if one diagonal bisects the area of the quadrilateral.

- 6 years, 2 months ago

Can you please provide the proof sir ?

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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

- 6 years, 2 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.

- 6 years, 2 months ago

Yeah , thats why I posted this as a note.

- 6 years, 2 months ago

Did you get a proof

- 6 years, 2 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.

- 6 years, 2 months ago

yes,same

- 6 years, 2 months ago

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

- 6 years, 2 months ago

Oh yes, thank you for pointing that out to me

- 6 years, 2 months ago

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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

- 6 years, 2 months ago

Yeah, Thanks! :) :D :)

- 6 years, 2 months ago

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

- 6 years, 2 months ago