Three, (Full marks 30 points) In $\triangle A B C$, $A B=A C$, the altitude $A D=5$ on $B C$, $M$ is a point on $A D$, $M D=1$, and $\angle B M C=3 \angle B A C$. Try to find the perimeter of $\triangle A B C$.
Three, draw $M E$ $/ / A B$ intersecting $B C$ at $E$, and draw $M F / / A C$ intersecting $B C$ at $F$. Let $E D=t, M E=$ $s$.
$$
\because \triangle B D A \backsim
$$
$\triangle E D M, \therefore \frac{B D}{E D}=\frac{A D}{M D}=\frac{5}{1}, \therefore B D=5 t$.
Also, $M E$ is the angle bisector of $\angle B M F$,
$$
\begin{array}{l}
\therefore \frac{B M}{M F}=\frac{B E}{E F}=\frac{4 t}{2 t}=\frac{2}{1}, \\
\therefore B M=2 M F=2 M E=2 s .
\end{array}
$$

By the Pythagorean theorem, we have
$$
\begin{array}{l}
B M^{2}-B D^{2}=4 s^{2}-25 t^{2}=1, \\
M E^{2}-E D^{2}=s^{2}-t^{2}=1,
\end{array}
$$

Solving (1) and (2) gives $t=\frac{\sqrt{7}}{7}$, thus
$$
s=\sqrt{1+t^{2}}=\frac{2 \sqrt{14}}{7} \text {. }
$$
$\therefore$ The perimeter of $\triangle A B C = 5 \cdot$ the perimeter of $\triangle M E F$
$$
\begin{array}{l}
=10(s+t)=\frac{10}{7}(\sqrt{7}+2 \sqrt{14}) \\
=\frac{10 \sqrt{7}}{7}(1+2 \sqrt{2}) .
\end{array}
$$