[Geometry]
Problem: Let $A B C D$ be a parallelogram with $\angle A B C$ obtuse. Let $\overline{B E}$ be the altitude to side $\overline{A D}$ of $\triangle A B D$. Let $X$ be the point of intersection of $\overline{A C}$ and $\overline{B E}$, and let $F$ be the point of intersection of $\overline{A B}$ and $\overleftrightarrow{D X}$. If $B C=30, C D=13$, and $B E=12$, compute the ratio $\frac{A C}{A F}$.

Solution: Extend $\overline{A D}$ to a point $M$ such that $\overline{C M} \| \overline{B E}$ as shown below.

<img_3958>

Because $C D=A B=13$ and $B E=12=C M, A E=D M=5$. Then $A C=\sqrt{35^{2}+12^{2}}=$ $\sqrt{1369}=37$. Because $\overline{E X} \| \overline{C M}, X E / C M=A E / A M=\frac{1}{7}$. Thus $E X=\frac{12}{7}$ and $X B=\frac{72}{7}$, from which $E X / X B=\frac{1}{6}$. Apply Menelaus's Theorem to $\triangle A E B$ and Menelaus line $\overline{F D}$ :

$$
\begin{aligned}
\frac{A D}{E D} \cdot \frac{E X}{X B} \cdot \frac{B F}{F A} & =1 \\
\frac{30}{25} \cdot \frac{1}{6} \cdot \frac{13-F A}{F A} & =1 \\
\frac{13-F A}{F A} & =5 .
\end{aligned}
$$

Thus $F A=\frac{13}{6}$. The desired ratio is:

$$
\frac{37}{13 / 6}=\frac{\mathbf{2 2 2}}{\mathbf{1 3}}
$$
After calculating $A C$ as above, draw $\overline{B D}$, intersecting $\overline{A C}$ at $Y$. Because the diagonals of a parallelogram bisect each other, $D Y=Y B$. Then apply Ceva's Theorem to $\triangle A B D$ and concurrent cevians $\overline{A Y}, \overline{B E}, \overline{D F}$ :

$$
\begin{aligned}
& \frac{A E}{E D} \cdot \frac{D Y}{Y B} \cdot \frac{B F}{F A}=1 \\
& \frac{5}{25} \cdot 1 \cdot \frac{13-F A}{F A}=1
\end{aligned}
$$

Thus $F A=\frac{13}{6}$, and the desired ratio is $\frac{\mathbf{2 2 2}}{\mathbf{1 3}}$.
By AA similarity, note that $\triangle A F X \sim \triangle C D X$ and $\triangle A E X \sim \triangle C B X$. Thus $\frac{A F}{C D}=\frac{A X}{X C}=\frac{A E}{C B}$. Thus $\frac{A F}{13}=\frac{A E}{C B}=\frac{5}{30}$, so $A F=\frac{13}{6}$, and the answer follows after calculating $A C$, as in the first solution.

Final answer: $\frac{222}{13}$