Let $K$ a point on the segment $AB$ s.t $KB=DC$ since $CM$ bisects the area of the trapezium then it 's clear that $M$ is the midpoint of $AK$ ,Let $d\equiv AM,d'\equiv DC ,h$ the height of $ A$ from $DC$and $h',h''$ the heights of N from $AB,DC$ resp..since $AN$ bisects the area of the trapezium then $(2d+d')h'=(d+d')h\implies \frac{2d+d'}{h}=\frac{d+d'}{h'}=\frac{d}{h''}\implies \frac{h'}{h''}=\frac{d+d'}{d}$ thus $\frac{NB}{NC}=\frac{d+d'}{d}$ ; consider $G$ the intersection of $MN$ and $CK$ ,applying Menelaus to the triangle $CKB$ and the points $ M-G-N$ yields $ \frac{NB}{NC}\cdot \frac{MK}{MG}\cdot \frac{GC}{GK}=1\implies \frac{d+d'}{d}\cdot \frac{d}{d+d'}\cdot \frac{GC}{GK}=1$ thus $G$ is the midpoint of $CK$ but $KBCD$ is a parallelogram hence $G$ is the midpoint of $BD$. R HAS

Let h be the trapezium’s height. (AM + CD)h/2 = (BM)h/2 AM + CD = BM. Let [XYZ] denote area of a polygon XYZ. [CBM] = [ANB] [BMN] + [CMN] = [BMN] + [AMN] [CMN] = [AMN] Therefore, MN//AC (i.e. because triangles between parallel lines with common base have equal areas and vice versa). Let MN intersect DC at N’ Hence AMN’C is a parallelogram because MN//AC therefore MN’//AC. Hence CN’ = AM. AM + CD = CN’ + CD = BM Hence BM = DN’. BM//DC//DN ’and BM = DN’, hence BMDN’ is a parallelogram. In a parallelogram the diagonals bisect each other at the point of intersection. Hence MN’ and MN pass through the midpoint of BD. Q.E.D.

It can be done analitically in 5 minutes. It's ugly so I won't post the solution.

I am in middle school AND what is a cyclic series the sigma with cyc on the bottom and what does it mean.

LOCUS wrote: I am in middle school AND what is a cyclic series the sigma with cyc on the bottom and what does it mean. For example, given $3$ variables $a, b, c$, $$\sum_{cyc} a^2b = a^2b + b^2c + c^2a$$ A more rigorous definition can be found here.

Mmesomachi wrote: Let $h$ be the trapezium’s height. Since $(AM + CD) \cdot \frac{h}{2} = (BM) \cdot \frac{h}{2}$, we have that $AM + CD = BM$. Let $[XYZ]$ denote area of a polygon $XYZ$. $[CBM] = [ANB] \Rightarrow [BMN] + [CMN] = [BMN] + [AMN] \Rightarrow [CMN] = [AMN]$ Therefore, $MN \parallel AC$ (i.e. because triangles between parallel lines with common base have equal areas and vice versa). Let $MN$ intersect $DC$ at $N’$. Hence $AMN’C$ is a parallelogram because $MN \parallel AC$ therefore $MN’ \parallel AC$. Hence $CN’ = AM$. Then, $AM + CD = CN’ + CD = BM$ Hence $BM = DN’$. $BM \parallel DC \parallel DN’$ and $BM = DN’$, hence $BMDN’$ is a parallelogram. In a parallelogram the diagonals bisect each other at the point of intersection. Hence $MN’$ and $MN$ pass through the midpoint of $BD$.$\square$