13.55 As shown in the figure, in triangle $A B C$, $A D$ and $A E$ bisect $\angle B A C$, and the lengths of $B D$, $D E$, and $E C$ are 2, 3, and 6, respectively. What is the length of the shortest side of $\triangle A B C$?
(A) $2 \sqrt{10}$.
(B) 11 .
(C) $6 \sqrt{6}$.
(D) 6 .
(E) Cannot be uniquely determined by the given conditions.
[Solution 1] As shown in the figure, let $\angle BAC = 3\alpha$, $x = AB$, and $y = AD$. By the angle bisector property, we have
$$
\frac{AB}{AE} = \frac{2}{3} \quad \text{and} \quad \frac{AD}{AC} = \frac{1}{2},
$$
thus $AE = \frac{3}{2}x$ and $AC = 2y$.

Applying the cosine rule in $\triangle ADB$, $\triangle AED$, and $\triangle ACE$, we get
$$
\left\{\begin{array}{l}
4 = x^2 + y^2 - 2xy \cos \alpha, \\
9 = \frac{9}{4}x^2 + y^2 - 3xy \cos \alpha, \\
36 = \frac{9}{4}x^2 + 4y^2 - 6xy \cos \alpha.
\end{array}\right.
$$

Thus, we have
$$
\frac{x^2 + y^2 - 4}{2xy} = \frac{\frac{9}{4}x^2 + y^2 - 9}{3xy},
$$

and
$$
\frac{x^2 + y^2 - 4}{2xy} = \frac{\frac{9}{4}x^2 + 4y^2 - 36}{6xy}.
$$

From (1), we get
$$
3x^2 - 2y^2 = 12,
$$

From (2), we get
$$
3x^2 - 4y^2 = -96.
$$

Solving these two equations, we obtain $x = 2\sqrt{10}$ and $y = \sqrt{54} = 3\sqrt{6}$. Therefore, $AB = 2\sqrt{10} \approx 6.3$, $AC = 6\sqrt{6} \approx 14.7$, and $BC = 11$. Hence, the answer is $(A)$.

[Solution 2] As shown in the figure, let $AB = a$, $AD = b$, $AE = c$, and $AC = d$. By the angle bisector property, we have
$$
\frac{a}{c} = \frac{2}{3}, \quad \frac{b}{d} = \frac{3}{6},
$$

thus $a = \frac{2}{3}c$ and $d = 2b$.
By the angle bisector length formula, we have
$$
b^2 + 6 = ac \text{ and } c^2 + 18 = bd,
$$

From the above, we get $b^2 + 6 = \frac{2}{3}c^2$ or $c^2 + 18 = 2b^2$, solving these gives
$$
c^2 = 90, \quad b^2 = 54.
$$
Therefore, $a = \frac{2(3\sqrt{10})}{3} = 2\sqrt{10}$, and $d = 2(3\sqrt{6}) = 6\sqrt{6}$.
Hence, the answer is $(A)$.