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?
 A: 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.
A: Area $GBCM = GB \cdot BC$ ; Area $GBEF = GB \cdot \frac12(GF+BE) = GB \cdot RN$.
And Area $GBEF = 2 \cdot$ Area $FECM$, so Area $GBEF = \frac23$ Area $GBCM$. Hence $RN = \frac23 BC$.
A: First using similarity of the triangles $\Delta ABE$ and $\Delta ARN$ you get $\dfrac{RN}{BE}=\dfrac{5}{6}$.
Then using the given ratio of areas of $GBEF$ and $FECM$,
$$\frac{BE+GF}{2}=EC+FM=2BC-BE-GF\implies BE+GF={4 \over 3}BC$$
Using similarity of the triangles $\Delta AGF$ and $\Delta ABE$ you get $\dfrac{GF}{BE}=\dfrac{2}{3}\implies BE=\dfrac{4}{5}BC$
Can you take it from here?
