My solution: Let $ X. Y $ be the midpoint of $ EF, AK $ . Easy to see $ M $ is the common midpoint of $ AN $ and $ EL $ . Since $ X, M $ is the midpoint of $ EF, EL $ so $ MX\parallel LF $ . ... $ (1) $ Since $ Y, M $ is the midpoint of $ AK, AN $ so $ MY\parallel NK $ . ... $ (2) $ From Newton line theorem we get $ M, X, Y $ are collinear , so combine with $ (1) $ and $ (2) $ we get $ LF\parallel NK $ . Q.E.D

hajimbrak wrote: Let $M$ be the midpoint of the side $BC$ of $\triangle ABC$. On the side $AB$ and $AC$ the points $\color{red} E$ and $\color{red} F$ are chosen. Let $K$ be the point of the intersection of $BF$ and $CE$ and $L$ be chosen in a way that $CL\parallel AB$ and $BL\parallel CE$. Let $N$ be the point of intersection of $AM$ and $CL$. Show that $KN$ is parallel to $FL$. Typo fixed. $ABNC$ and $EBLC$ are parallelograms $\implies$ intersections $X_\infty \equiv AB \cap NC \equiv EB \cap LC,$ $Y_\infty \equiv BN \cap CA$ and $Z_\infty \equiv BL \cap CE$ of their opposite side pairs are all at infinity. $\exists$ central projection, taking common diagonal $BC$ of these parallelograms to the line at infinity $B'_\infty C'_\infty.$ It also takes the previous line at infinity $X_\infty Y_\infty Z_\infty$ to a finite line $X'Y'Z'.$ All projected points are labeled with primes. Projected quadrilaterals $A'X'N'Y'$ and $E'X'L'Y'$ are parallelograms. Let $P' \equiv N'K' \cap A'Y',$ $Q' \equiv N'P' \cap F'X'$ and $S' \equiv X'Y' \cap F'N'.$ From centrally similar $\triangle S'N'X' \sim \triangle S'F'Y'$ and $\triangle Q'X'N' \sim \triangle Q'F'P'$ $\implies$ $\frac{\overline{S'N'}}{\overline{S'F'}} = \frac{\overline{N'X'}}{\overline{F'Y'}}$ and $\frac{\overline{Q'F'}}{\overline{Q'X'}} = \frac{\overline{F'P'}}{\overline{X'N'}},$ resp. It follows that $\frac{\overline{S'N'}}{\overline{S'F'}} \cdot \frac{\overline{Q'F'}}{\overline{Q'X'}} \cdot \frac{\overline{F'Y'}}{\overline{F'P'}} = -1.$ Let $D' \equiv E'Z' \cap N'Y'.$ From centrally similar $\triangle N'Y'P' \sim \triangle K'F'P'$ and $\triangle Y'X'N' \sim \triangle Y'Z'D'$ $\implies$ $\frac{\overline{Y'P'}}{\overline{F'P'}} = \frac{\overline{N'Y'}}{\overline{K'F'}} = \frac{\overline{N'Y'}}{\overline{D'Y'}} = \frac{\overline{X'N'}}{\overline{Z'D'}} = \frac{\overline{X'N'}}{\overline{L'N'}}$ $\implies$ $\frac{\overline{F'Y'}}{\overline{F'P'}} = \frac{\overline{L'X'}}{\overline{L'N'}}$ and $\frac{\overline{S'N'}}{\overline{S'F'}} \cdot \frac{\overline{Q'F'}}{\overline{Q'X'}} \cdot \frac{\overline{L'X'}}{\overline{L'N'}} = -1.$ By Ceva theorem for $\triangle X'N'F'$ $\implies$ lines $X'S'Y', N'Q'P', F'L'$ are concurrent at a point $T'.$ (Note that $K' \in N'Q'P'.$) Since $T' \in X'Y'Z'$ $\implies$ the original point $T \equiv KN \cap FL$ before central projection is on the line at infinity, $T \equiv T_\infty \in X_\infty Y_\infty Z_\infty,$ making the lines $KN \parallel FL$ parallel.

Attachments:

Solution using Pappus line.

Attachments:

DK/DF=DK/DB.DB/DF=CK/CE.CA/CF=(FA/FC+1)/(KE/KC+1)=(AB/CD+1)/(BE/CD+1)=AB+CD/BE+CD=DN/DL ->q.e.d

Attachments: