TITLE: Prove that $MN$ is parallel to $KL$.
QUESTION [6 upvotes]: In triangle $ABC$,$E$ and $F$ are points on the sides $AB$ and $AC$ .We name the intersection of $BF$ and $CE$, point $D$.We choose the points $K$,$L$,$M$ and $N$ on sides $AC$,$CE$,$AB$ and $BF$ that $BM=EA$,$BN=FD$,$CL=ED$ and $KC=AF$.Prove that $MN$ is parallel to $KL$.

It seems that similarity will work but it didn't.And we don't have any angle information to show the lines are parallel then how should I work?

REPLY [4 votes]: The vectors will do:
\begin{eqnarray*}
  \overrightarrow{NM} &=& \overrightarrow{NB}+\overrightarrow{BM}\\
   &=& \overrightarrow{FD}+\overrightarrow{EA}\\
   &=& \overrightarrow{D}-\overrightarrow{F}+\overrightarrow{A}-\overrightarrow{E}
\end{eqnarray*}
\begin{eqnarray*}
  \overrightarrow{LK} &=& \overrightarrow{LC} + \overrightarrow{CK}\\
  &=& \overrightarrow{ED} + \overrightarrow{FA}\\
   &=& \overrightarrow{D}-\overrightarrow{E}+\overrightarrow{A}-\overrightarrow{F}
\end{eqnarray*}
Since $\overrightarrow{NM}=\overrightarrow{LK}$ the quadrilateral
$MNLK$ is a parallelogram.