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 Fischer Aug 22 '18 at 9:00
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Please take a look at problem 21 in this pdf.
- 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.
- 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.