In the figure below, $A B C D$ and $A E F G$ are rectangles and the point $F$ lies on the diagonal $A C$. The area of the gray triangle is equal to $\frac{1}{18}$ of the area of the rectangle $A E F G$. What is the value of $\frac{A F}{A C}$?

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-15.jpg?height=317&width=511&top_left_y=1343&top_left_x=681)
A) $\frac{3}{5}$
B) $\frac{3}{8}$
C) $\frac{8}{13}$
D) $\frac{11}{18}$
E) $\frac{3}{4}$

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Solution

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-47.jpg?height=365&width=603&top_left_y=1425&top_left_x=635)

Since the area of triangle RFS is equal to $\frac{1}{18}$ of the area of rectangle $A E F G$, it is equal to $\frac{1}{9}$ of the area of triangle $E F G$. Since these triangles are similar and the ratio of their areas is the square of their similarity ratio, it follows that this latter ratio is $\sqrt{\frac{1}{9}}=\frac{1}{3}$. Thus, $F R=\frac{1}{3} E F$ and then $E R=E F-\frac{1}{3} E F=\frac{2}{3} E F$. Since triangles $F R S$ and $E B R$ are similar, this shows that their similarity ratio is

$$
\frac{F R}{R E}=\frac{\frac{1}{3} E F}{\frac{2}{3} E F}=\frac{1}{2}
$$

We then have $A E=G F=3 F S$ and $E B=2 F S$, so $A B=A E+E B=3 F S+2 F S=5 F S$ and $\frac{A E}{A B}=\frac{3 F S}{5 F S}=\frac{3}{5}$. By the theorem of Thales, we have $\frac{A F}{A C}=\frac{A E}{A B}$ and we obtain $\frac{A F}{A C}=\frac{3}{5}$.