How is $RN =\frac{2}{3}BC$ in this rectangle? 
I tried this question first by letting $AB = 6p$ and $BC = 6q$. Then I let $R$ the midpoint of $GB$ and $N$ the midpoint of $FE$. I got stuck so I looked at the memo for a hint and it turns out I followed the memo exactly but the next step said that $RN = \frac{2}{3}BC$. How is this true? I couldn't get that into my head. Can someone please explain?

We are free to assume that $ABCD$ is the unit square. 
Then assuming $BE=x$ we have $GF=\frac{2}{3}x$ by Thales' theorem and we must have:
$$\frac{GF+BE}{MF+CE}=\frac{\frac{2}{3}x+x}{1-\frac{2}{3}x+1-x}=2,$$
from which it follows that $x=\frac{4}{5}$. That implies $IH=\frac{4}{15}$. Since $AI=\frac{1}{3}$,
$$\frac{[AIH]}{[ABC]}=\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{15}=\frac{2}{45},$$
hence $(C)$ is the right answer. Since $RN=\frac{GF+BE}{2}$, it follows that $RN=\frac{5}{6}x=\frac{2}{3}$ as wanted.
