AMC 12 Mock Exam 2 Problem 23

By the angle bisector theorem, $BD=\frac{7}{15}\cdot9=\frac{21}5$ and $CD=\frac{8}{15}\cdot9=\frac{24}5$.

Line $BC$ is tangent to the circle at $D$, so the power of $B$ gives $BD^2=BE\cdot BA$, i.e., $BE=\frac{441/25}{7}=\frac{63}{25}$ and $AE=7-\frac{63}{25}=\frac{112}{25}$. Similarly $CD^2=CF\cdot CA$ gives $CF=\frac{576/25}{8}=\frac{72}{25}$ and $AF=8-\frac{72}{25}=\frac{128}{25}$. (Both $E$ and $F$ lie on the sides, since $BE<BA$ and $CF<CA$.)

Thus $\frac{AE}{AB}=\frac{112}{175}=\frac{16}{25}$ and $\frac{AF}{AC}=\frac{128}{200}=\frac{16}{25}$, so $\triangle AEF\sim\triangle ABC$ (SAS) with ratio $\frac{16}{25}$, and $EF=\frac{16}{25}\cdot9=\boxed{\tfrac{144}{25}}$.

(In general $\frac{AE}{AB}=1-\frac{BD^2}{AB^2}=1-\left(\frac{a}{b+c}\right)^2=\frac{AF}{AC}$, so $\overline{EF}\parallel\overline{BC}$ for every triangle.)