19. Let $\triangle A B C$ be an equilateral triangle with side length 16 . Points $D, E, F$ are on $C A, A B$, and $B C$, respectively, such that $D E \perp A E, D F \perp C F$, and $B D=14$. The perimeter of $\triangle B E F$ can be written in the form $a+b \sqrt{2}+c \sqrt{3}+d \sqrt{6}$, where $a, b, c$, and $d$ are integers. Find $a+b+c+d$.
19. 31
Solution. Let $\overline{A D}=2 x$, then $\overline{D C}=16-2 x$. Since $\triangle D A E$ and $\triangle D C F$ are both $30-60-90$ triangles, then $\overline{A E}=\overline{A D} / 2=x, \overline{E D}=x \sqrt{3}$ and $\overline{C F}=\overline{D C} / 2=8-x, \overline{F D}=(8-x) \sqrt{3}$. Since $\overline{A B}=\overline{B C}=16$, then $\overline{E B}=16-x$ and $\overline{F B}=8+x$. Using Pythagorean Theorem on $\triangle D E B$, we have
$$
\begin{aligned}
\overline{E D}^{2}+\overline{E B}^{2} & =\overline{B D}^{2} \\
(x \sqrt{3})^{2}+(16-x)^{2} & =14^{2} \\
3 x^{2}+256-32 x+x^{2} & =196 \\
x^{2}-8 x+15 & =0 \\
(x-5)(x-3) & =0 \\
x & =5,3
\end{aligned}
$$

Choosing either of the two values of $x$ will give the same result for the perimeter of $\triangle B E F$. Suppose we choose $x=5$, then $\overline{E B}=11$ and $\overline{F B}=13$. By Cosine Law on $\triangle B E F$, we have
$$
\begin{aligned}
\overline{E F}^{2} & =\overline{E B}^{2}+\overline{F B}^{2}-2 \cdot \overline{E B} \cdot \overline{F B} \cos 60^{\circ} \\
\overline{E F}^{2} & =11^{2}+13^{2}-2(11)(13)(1 / 2) \\
\overline{E F} & =\sqrt{121+169-143}=7 \sqrt{3}
\end{aligned}
$$

Thus, the perimeter of $\triangle B E F=24+7 \sqrt{3}$, which means that $a+b+c+d=24+0+7+0=31$.