Here is a 9th grade problem from the 1957 Moscow Mathematical Olympiad. The problem was also included into a problem collection by V. V. Prasolov

In quadrilateral $ABCD$, $M$ is the midpoint of the diagonal $AC$, $N$ the midpoint of $BD$. The straight line through $M$ and $N$ meets $AB$ and $CD$ in points $M'$ and $N'$. Prove that, if $MM' = NN'$, then $AD\parallel BC$.

19 January 2015, Created with GeoGebra

### Proof

For a proof, we may assume that $M\ne N$, for, otherwise, the two diagonals will be halved by their point of intersection, making the quadrilateral $ABCD$ a parallelogram so that obviously $AD\parallel BC$.

Let $E$ and $F$ be the midpoints of $CD$ and $AB$, respectively.

If $MN\parallel BC$ then $F=M'$, $E=N'$, and $MN\parallel AD$ and we are done. So assume $MN\not\parallel BC$. Then $F\ne M'$ and $E\ne N'$. Define $K$ to be the midpoint of $BC$.

There are now several ways to proceed. For example, one depends on the fact that $MN$ and $EF$ meet in point $O$ which divides both in half. Below I chose the least sophisticated endgame.

Now, $\Delta FMM' = \Delta ENN'$. Indeed, $\displaystyle FM=\frac{BC}{2}=EN$ as midlines in triangles $ABC$ and $BCD$; $MM'=NN'$ as a problem premise, and $\angle FMM'=\angle ENN'$, because $FM\parallel BC\parallel EN$. If so, $\angle AFM=\angle CEN$ such that $FM'\parallel EN'$ but then also $MK\parallel NK$, leading to a contradiction ($M=N$.)

### References

1. V. V. Prasolov, Problems in Planimetry, v 1, Nauka, Moscow, 1986 (Russian), 12.8