Square $ABCD$ has side length $12$. Point $E$ lies on $\overline{AB}$ with $AE=9$ and point $F$ lies on $\overline{BC}$ with $BF=9$. Segments $\overline{DE}$ and $\overline{AF}$ meet at $G$. What is the area of quadrilateral $EBFG$?

1. Right triangles $ADE$ and $BAF$ are congruent ($AD=BA=12$, $AE=BF=9$), so $\angle ADE=\angle BAF$. Hence $\angle DAG+\angle ADG=\angle DAG+\angle BAF=90^\circ$, so $\angle AGD=90^\circ$ and $AF\perp DE$.
2. In right triangle $ADE$, $DE=\sqrt{12^2+9^2}=15$. The altitude $AG$ to $DE$ gives $AG=\frac{12\cdot9}{15}=\frac{36}5$ and $GE=\frac{AE^2}{DE}=\frac{81}{15}=\frac{27}5$.
3. $[EBFG]=[ABF]-[AGE]=\frac12\cdot12\cdot9-\frac12\cdot\frac{36}5\cdot\frac{27}5=54-\frac{486}{25}=\boxed{\frac{864}{25}}$.