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$
Solution
Reflect $B$ in $AC$ to $B'$ and $C$ in $AB$ to $C'$. Then $BE=B'E$ and $CD=C'D$, so \[BE+DE+CD=B'E+ED+DC'\ge B'C'.\]
$AB'=10$, $AC'=6$, $\angle B'AC'=3\cdot40^\circ=120^\circ$, so $B'C'^2=10^2+6^2-2\cdot10\cdot6\cos120^\circ=196$, $B'C'=14$.
$B'C'$ meets $AC$ and $AB$ at points $E,D$ on the sides (order $B',E,D,C'$), so equality holds.
In $\triangle BAC$, $\angle BAC=30^\circ$, $AB=12$, $AC=8$. Points $D,E$ lie on $\overline{AB},\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$?
