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Exercise 382
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 $BE+DE+CD$?
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Solution
  1. Reflect $B$ in $AC$ to $B'$ and $C$ in $AB$ to $C'$. Then $BE+DE+CD=B'E+ED+DC'\ge B'C'$.
  2. $AB'=12$, $AC'=8$, $\angle B'AC'=3\cdot30^\circ=90^\circ$, so $B'C'^2=12^2+8^2=208$.
  3. $B'C'$ meets $AC$, $AB$ at points $E,D$ on the sides, so equality holds.
  4. The minimum is $\sqrt{208}=\boxed{4\sqrt{13}}$.
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