If $\dfrac{AM}{MB}=\dfrac{m}{n},$ then why $AM=\dfrac{m}{m+n}AB?$
I am trying to show that if $M$ lies on $AB$ and $AM:MB=m:n$, then $$\vec{OM}=\dfrac{n\vec{OA}+m\vec{OB}}{m+n}.$$ For that problem I am supposed to notice $$AM=\dfrac{m}{m+n}$$ but I don't see it.