AMC 12 2014a Problem 20

In $\triangle BAC$, $\angle BAC=40^\circ$, $AB=10$, and $AC=6$. Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$?

$\textbf{(A) }6\sqrt 3+3 \textbf{(B) }\dfrac{27}2 \textbf{(C) }8\sqrt 3 \textbf{(D) }14 \textbf{(E) }3\sqrt 3+9$