2013 AMC 10B Problem 23

Let $BD=x$, so $CD=14-x$. Applying the Pythagorean Theorem to right triangles $ABD$ and $ACD$ gives $$13^2-x^2=15^2-(14-x)^2.$$ Thus $BD=5$, $CD=9$, and $AD=12$. This yields the following diagram: Because $AD\perp BC$ and $DE\perp AC$, we have $\angle ADE=\angle ACD$. Also, $\angle AFB=\angle ADB=90^\circ$, so $A,B,D,F$ are concyclic. Hence $$\angle ABF=\angle ADF=\angle ACD.$$ In right triangle $ACD$, $\cos(\angle ACD)=\frac35$ and $\sin(\angle ACD)=\frac45$. Therefore, in right triangle $ABF$, $$\begin{aligned} BF&=13\cdot\frac35,\\ AF&=13\cdot\frac45. \end{aligned}$$ By Ptolemy’s Theorem, we get $$\begin{aligned} &AB \cdot DF + DB \cdot AF \\ &= BF\cdot AD . \end{aligned}$$ Therefore, $$\begin{aligned} 13\cdot DF&+5\left(13\cdot\frac45\right)\\ &=12\left(13\cdot\frac35\right). \end{aligned}$$ Dividing by $13$ gives $DF+4=\frac{36}{5}$, so $DF=\frac{16}{5}$. Thus $m+n=21$, and the correct answer is B .