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Let $ABCD$ be a convex quadrilateral and let $O$ be the point of intersection of its diagonals. Prove that if the perimeters of $\triangle ABO$,$\triangle BCO$,$\triangle CDO$ and $\triangle DAO$ are equal then $ABCD$ is a rhombus.

first time I saw this problem on Problem-Solving Strategies book By Arthur Engel on summer(you can see this solution from this book on page 126).this problem was in the extremal principle chapter.I wanted to find another way to prove it but I was not able to do anything useful.

I did some research and I was able to find that this problem is From MOSP 2001. But I was not able to find any other solution other than the one I saw on Arthur Engel book. Any help or hint on proving this problem in other ways are appreciated.

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  • $\begingroup$ Can you tell us what that solution was like? $\endgroup$ – Sawarnik Nov 20 '14 at 15:38
  • $\begingroup$ @Sawarnik what do you mean from "that" solution? the only solution i found for this problem is the one that I mentioned in my post. $\endgroup$ – user2838619 Nov 20 '14 at 17:53
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We see that $ OB + BA = OD + DA $, hence $B, D$ lie on an ellipse whose foci is $A, O$.
Similarly, since $OB + BC = OD + DC $, hence $B, D$ lie on an ellipse whose foci is $ O, C $.

Slight argument that ellipse with foci $A,O$ and $O,C$, with $AOC$ a straight line can only intersect at 2 points. Furthermore, by symmetry, these 2 points are symmetric about the line $AOC$. Thus $BD \perp AC$.

From this, it follows that $ AB = BC = CD = DA$ from the perimeter requirements, hence we have a rhombus.

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