Let $C_1$ be the reflection of $C$ across $\overline{AB}$, and let $C_2$ be the reflection of $C_1$ across $\overline{AC}$.  Then it is well-known that the quantity $BE+DE+CD$ is minimized when it is equal to $C_2B$.  (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.)  As $A$ lies on both $AB$ and $AC$, we have $C_2A=C_1A=CA=6$.  Furthermore, $\angle CAC_1=2\angle CAB=80^\circ$ by the nature of the reflection, so $\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ$.  Therefore by the Law of Cosines \[BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{\textbf{(D) }14}.\]
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 $BE+DE+CD$?
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$