# Prove that $MN$ is parallel to $KL$.

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?

• @JackD'Aurizio I wrote them t:Try proving similarity or using angles. Apr 22, 2017 at 16:32

You can construct point $P$ so that $PF$ is parallel to $AB$ and $PE$ is parallel to $AC$. Then $AEPF$ is a parallelogram and then one can easily see that triangle $FPD$ is congruent to triangle $BMN$ as well as triangle $EPD$ is congruent to triangle $CKL$. Consequently, $\angle \, PDN = \angle \, MND$ and $\angle \, PDL = \angle \, KLD$ so $MN$ is parallel and equal to $PD$ and $KL$ is parallel and equal to $PD$. Thus $MN$ is parallel and equal to $KL$.
The other solutions is similar, but you construct the point $Q$ instead. Again you end up with congruent triangles and analogous arguments. Indeed, construct point $Q$ so that $QE$ is parallel to $BF$ and $QF$ is parallel to $CE$. Then $DEQF$ is a parallelogram and then one can easily see that triangle $EAQ$ is congruent to triangle $BMN$ as well as triangle $FAQ$ is congruent to triangle $CKL$. Consequently, $\angle \, BMN = \angle \, EAQ$ and $\angle \, CKL = \angle \, FAQ$ so $MN$ is parallel and equal to $AQ$ and $KL$ is parallel and equal to $AQ$. Thus $MN$ is parallel and equal to $KL$.
\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.
Since affine maps preserve midpoints and parallel lines, it is not restrictive to assume that $A$ lies at the origin, $B$ has coordinates $(1,0)$, $C$ has coordinates $(0,1)$, $F$ has coordinates $(0,c)$, $E$ has coordinates $(b,0)$. Find the coordinates of $D,M,K,N,L$ and the claim will be straightforward to prove. An equivalent approach is to use barycentric coordinates.