$O$ is the point inside triangle $ABC$. The lines joining the three vertices $A, B, C$ to $O$ cut the opposite sides in $K, L,$ and $M$ respectively.
A line through $M$ parallel to $KL$ cuts the line $BC$ at $V$ and $AK$ at $W$.
Prove that $VM$=$MW$.
I tried using the Ceva's theorem in triangle $ABC$ and equating it with the relation obtained by Menelaus's theorem. Then I used basic proportionality theorem and after using them I just got one relation. $$\frac{BK}{BV}=\frac{AL}{AP}$$
Now I am stuck please help me.
Solution by ratios
Let $AB$ intersect $KL$ at $X$. The result follows by multiplying these five equations and cancelling:
$$\begin{align*} \frac{VM}{MW} \cdot \frac{WA}{AK} \cdot \frac{KB}{BV} &= -1 &&\text{(Menelaus)} \\[.5em] \frac{AX}{XB} \cdot \frac{BK}{KC} \cdot \frac{CL}{LA} &= -1 &&\text{(Menelaus)} \\[.5em] 1 &= \frac{AM}{MB} \cdot \frac{BK}{KC} \cdot \frac{CL}{LA} &&\text{(Ceva)} \\[.5em] \frac{AM}{AW} &= \frac{AX}{AK} &&\text{(Thales)} \\[.5em] \frac{BV}{BM} &= \frac{BK}{BX} &&\text{(Thales)} \\[1em] \hline \frac{VW}{MW} &= 1 \end{align*}$$
Solution by projective geometry
Let $AB$ intersect $KL$ at $X$. The complete quadrangle $OKCL$ shows that $M, X$ are harmonic conjugates with respect to $A, B$. Since a perspectivity through $K$ preserves harmonic conjugacy, $M, ∞$ are harmonic conjugates with respect to $W, V$; that is, $VM = MW$.