Locus of intersection point of two lines

If you wish to avoid classical geometry, you can use vector geometry for questions of this type.

Take the origin at $A$ and let $B,C$ have position vectors $b,c$ respectively. Let $L,K$ have position vectors $tc,sb$ respectively. $M$ is on the line $LK$ and so $$tc-\frac{b+c}{2}=\lambda \left(\frac{b+c}{2}-sb \right).$$

Equating coefficients of $b$ and $c$ gives the position vectors of $L,K$ as respectively $$ \frac{\lambda+1}{2}c,\frac{\lambda+1}{2\lambda}b.$$

By the ratio theorem any point on $BL$ has position vector $$rb+(1-r)\frac{\lambda+1}{2}c$$ for some $r$. Similarly, any point on $CK$ has position vector $$sc+(1-s)\frac{\lambda+1}{2\lambda}b$$ for some $s$. The point $N$ is where $BL$ and $CK$ intersect and equating the coefficients of $b$ and $c$ in the two expressions gives the position vector of $N$ as $$\frac{\lambda+1}{\lambda-1}(b-c).$$ Therefore the locus of $N$ is indeed a line throught the origin $A$ parallel to $BC$.

A more geometrical answer

Let the line through $M$ intersect $AB$ extended at $K$. Let the line $KC$ intersect the line through $A$ which is parallel to $BC$ at $N'$. (We will prove $N=N'$.)

Apply a shear, fixing $A$ and $N'$ and making $AK=KN'$. After the shear, $KML$ is the line of symmetry of triangle $AKN'$ and so $N=N'$. Therefore this was true before the shear!