9. Let $A B C D$ be a square of side length 13 . Let $E$ and $F$ be points on rays $A B$ and $A D$, respectively, so that the area of square $A B C D$ equals the area of triangle $A E F$. If $E F$ intersects $B C$ at $X$ and $B X=6$, determine $D F$.
Answer: $\sqrt{13}$

First Solution
Let $Y$ be the point of intersection of lines $E F$ and $C D$. Note that $[A B C D]=[A E F]$ implies that $[B E X]+[D Y F]=[C Y X]$. Since $\triangle B E X \sim \triangle C Y X \sim \triangle D Y F$, there exists some constant $r$ such that $[B E X]=r \cdot B X^{2},[Y D F]=r \cdot C X^{2}$, and $[C Y X]=r \cdot D F^{2}$. Hence $B X^{2}+D F^{2}=C X^{2}$, so $D F=\sqrt{C X^{2}-B X^{2}}=\sqrt{49-36}=\sqrt{1} 3$.
Second Solution
Let $x=D F$ and $y=Y D$. Since $\triangle B X E \sim \triangle C X Y \sim \triangle D F Y$, we have
$$
\frac{B E}{B X}=\frac{C Y}{C X}=\frac{D Y}{D F}=\frac{y}{x} .
$$

Using $B X=6, X C=7$ and $C Y=13-y$ we get $B E=\frac{6 y}{x}$ and $\frac{13-y}{7}=\frac{y}{x}$. Solving this last equation for $y$ gives $y=\frac{13 x}{x+7}$. Now $[A B C D]=[A E F]$ gives
$$
\begin{aligned}
169 & =\frac{1}{2} A E \cdot A F=\frac{1}{2}\left(13+\frac{6 y}{x}\right)(13+x) . \\
169 & =6 y+13 x+\frac{78 y}{x} \\
13 & =\frac{6 x}{x+7}+x+\frac{78}{x+7} \\
0 & =x^{2}-13 .
\end{aligned}
$$

Thus $x=\sqrt{13}$.