Prove that the segment $MN$ intersects the segment $BD$ at its own midpoint. Let $ABCD$ be a trapezium such that the side $AB$ and $CD$ are parallel and the side $AB$ is longer than the side $CD$. Let $M$ and $N$ be on segments $AB$ and $BC$ respectively,such that each of the segments $CM$ and $AN$ divides the trapezium in two parts of equal area.
Prove that the segment $MN$ intersects the segment $BD$ at its own midpoint.

I worked out that $BM=AM+CD$ and if you let the midpoint be $O$ and you have $MN$ meet $CD$ at $K$ then $OKD$ must be congruent to $OMB$ for $O$ to be the midpoint
But Im not sure if that helps or what to do next, solutions would be appreciated
Taken from the 2016 Pan African Math Olympiad http://pamo-official.org/problemes/PAMO_2016_Problems_En.pdf
 A: You already know that $M$ satisfies $BM=AM+CD$. Let $X$ and $Y$ be points on $CD$ such that $ABXY$ is a parallelogram with $AY$ parallel to $MD$. 
Then $MX$ and $BD$ bisect each other since they are diagonals of parallelogram $MBXD$. 
From parallelogram $MBXD$ we also see that $XC+CD=BM=AM+CD$ i.e. $XC=AM$. Therefore $BC$ is parallel to $YM$. Let $MX$ intersect $BC$ at $N^*$. Then, from triangle $MXY$ we see that $$\frac{N^*C}{MY}=\frac{XC}{XY} \implies \frac{BN^*}{BC}=\frac{BM}{AB}.$$
Then the ratio of the area of $ABN$ to the area of $ABCD$ is $$\frac{1}{2}AB.BN^*:BM.BC=1:2.$$
Therefore $N^*=N$ and the proof is complete. 
A: Locate N: Extend BC to X such that DX // AC. Then, N is the midpoint of BX.
Let AX cut CD at Y.

The above construction ensures 0.5[ABCD] = [NADC] = [NCS] + [CSAY] + [YAD] = [NCS] + [CSAY] + [YCX] = [AXN] = [ABN].
M (such that BM = MZ) is similarly constructed on AB to meet the given requirement. 
Note that [BMSN] + [AMS] = [ABN] = 0.5[ABCD] = [BCM] = [BMSN] + [SCN].  MN // AC is required to ensure  [AMS] = [SCN].
Result follows by applying intercept theorem to $\triangle ABD$.
