Quadrilaterals and figures

Question

$ABC$ is a triangle in which $L$ is the midpoint of $AB$ and $N$ is a point on $AC$ such that $AN =2CN$. A line thought $L$ parallel to $BN$ meets $AC$ at $M$ prove that $AM=CN$.

Answer 1

By construction the triangles $ALM$ and $ABN$ are similar and $2AL=AB$, so we have also $2AM=AN$ and, since $AN=2CN$ we have $2AM=2CN \Rightarrow AM=CN$