2011 AMC 12B Problem 20

Since $DE\parallel AC$ and $EF\parallel AB,$ we get $\angle BDE=\angle BAC=\angle EFC.$ By the Inscribed Angle Theorem, $\angle BXE=\angle BDE$ and $\angle EXC=\angle EFC,$ so $\angle BXE=\angle EXC.$ With $BE=EC,$ this forces $XB=XC.$ Also $\angle BXC=2\angle BAC.$ The chord formula in $\triangle BXC$ gives $BC=2XB\sin(\angle BAC),$ while the same formula in $\triangle ABC$ gives $BC=2R\sin(\angle BAC).$ Thus $XB=XC=R.$ The point at distance $R$ from both $B$ and $C$ on this side of $BC$ is the circumcenter of $\triangle ABC,$ so $XA=XB=XC=R.$ The area of the $13$-$14$-$15$ triangle is $84$ by Heron’s formula, so $$R=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{65}{8},$$ and $XA+XB+XC=3R=\dfrac{195}{8}.$ Thus, the correct answer is C.