In the parallelogram $ABCD$, $\angle BAD = 60$. $E$ is the midpoint of $BC$ and $F$ is the midpoint of $CD$. $BD$ intersects $AE$ at $M$ and $AF$ at $N$. If $AB = 15\, $cm and $AD = 8\,$cm, what is the length, in cm, of $MN$?
From the 2014 IMC
I made a point $L$ on $DC$ so that $\:\angle BLC=90\:$ and used that to workout $LC= 4 $ and BL =$4\sqrt3$ and $BD=13$.
Because $ABCD$ is a parallelogram, $BC || AD$, and the transversals $AE$ and $BD$ intersect at $M$; hence $\triangle BME \sim \triangle DMA$. Moreover, because $BE = EC = BC/2$, it follows that the ratio of similitude is $1:2$, and in particular, $DM = 2BM$. We can apply the same reasoning with $\triangle FND \sim \triangle ANB$, with ratio of similitude $1:2$ (since $FD = 2AB$). Therefore $BN = 2DN$. Now since $$BD = BM + MN + ND = 3BM = 3DN,$$ we conclude $BM = DN = MN$; that is to say, $AE$ and $AF$ trisect $BD$.
All that remains is to calculate $BD$ via trigonometry; e.g., the law of cosines: $$BD^2 = 8^2 + 15^2 - 2(8)(15) \cos 60^\circ = 169,$$ hence $BD = 13$ and $MN = BD/3 = 13/3$.