smc

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 $C_{2}B$. (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 $C_{2}A=C_{1}A=CA=6$. Furthermore, $\angle CA{C}_1=2\angle CAB=80^{\circ }$ by the nature of the reflection, so $\angle C_{2}AB=\angle C_{2}AC+\angle CAB=80^{\circ }+40^{\circ }=120^{\circ }$. Therefore by the Law of Cosines $$B{C}_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^{\circ }=196⟹B{C}_2=\boxed{14}.$$