Midline in Quadrilateral

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


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 (\(M=N\).)


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

Related material

  • Thales' Theorem
  • Midline in Triangle
  • Midline in Trapezoid
  • Bimedians in a Quadrilateral
  • Midline in Similar Triangles
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