# Necessary/sufficient conditions for certain line arrangement in hexagon [closed]

Are there any necessary or sufficient condition for when in a general convex hexagon, the lines formed by joining the midpoints of opposite sides are concurrent (intersect in a common point)?

## closed as off-topic by Xander Henderson, Did, user21820, Daniel FischerAug 22 '18 at 9:00

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Did, user21820, Daniel Fischer
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• This question lacks context and seem pretty unmotivated to me. Can you please edit your question to provide some more information? For example, what is the source of this problem? Is this an exercise from a class? a contest problem? Why are you interested in it and, more importantly, why should anyone else care? Have you made any attempts to solve the problem? Where are you stuck? Have you drawn a picture? – Xander Henderson Aug 22 '18 at 3:24

Please take a look at problem 21 in this pdf.

1. Let $ABCDEF$ be a convex hexagon. If the area of $ACE$ is equal to the area of $BDF$ then the lines joining the midpoints of opposite sides concur.

If I recall correctly, the equality of these areas is also a necessary condition.

The proof uses problem 10 from the same pdf.

1. Let $ABCD$ be a convex quadrilateral. Let $AC$ intersect $BD$ at $E$. Let $P$ lies inside $ABCD$ such that the area of $BCP$ is equal to the area of $DAP$. Prove that the midpoints of $AB$, $CD$ and $EP$ are collinear.

Now, let me sketch a proof of 21.

Let $K,L,M,N,O,P$ be the midpoints of $AB, BC, CD ,DE, EF, FA$, respectively. Let $KN$ intersect $LO$ at $Q$. Let $S$ be the midpoint of $EB$. We use notation $[\mathcal F]$ to denote the area of $\mathcal F$. Then $$[KLS] = \frac 14 [ACE] = \frac 14 [BDF] = [SNO].$$ Using 10. we get that the midpoints of $QS, KO, LN$ are collinear. Denote this points by $X,Y,Z$, respectively. Note that $PKSO$ and $MLSN$ are parallelograms, so $Y$ and $Z$ are midpoints of $SP$ and $SM$, respectively. The homothety centered at $S$ and ratio $2$ maps $X,Y,Z$ to $Q,P,M$. It follows that $PM$ passes through $Q$ and we are done.

So, everything boils down to prove the thesis of problem 10. To prove it, you can adapt an argument from this post.

• Interesting. I was just going to comment a conjecture that the answer was that "the alternating sum of the areas of triangles formed by pairs of adjacent edges vanishes". This turns out to be an embarrassingly-inelegant way of getting at the same idea as your solution. – Blue Aug 4 '18 at 7:17
• Pls can u give proof of this statement, – nimmy Aug 4 '18 at 8:19
• @dasarinagavijaykrishna I have edited my post. – timon92 Aug 4 '18 at 10:08