24. In $\triangle A B C$, it is known that $M$ is the midpoint of side $A C$, $C N$ bisects $\angle A C B$, and intersects side $A B$ at point $N$, $B M$ intersects $C N$ at point $X$, and $\triangle B X N$ is an equilateral triangle, $A C=2$. Then $B N^{2}=(\quad)$.
(A) $\frac{10-6 \sqrt{2}}{7}$
(B) $\frac{2}{9}$
(C) $\frac{5 \sqrt{2}-3 \sqrt{3}}{8}$
(D) $\frac{\sqrt{2}}{6}$
(E) $\frac{3 \sqrt{3}-4}{5}$
24. A.

By Menelaus' theorem, we have
$$
\frac{A N}{N B} \cdot \frac{B X}{X M} \cdot \frac{M C}{C A}=1 \Rightarrow A N=2 X M \text {. }
$$

Let $A N=2 a, N B=b$. Then $B X=X N=b, M X=a$.
In $\triangle A B M$, by the cosine rule, we get
$$
\begin{array}{l}
1=A M^{2}=A B^{2}+B M^{2}-2 A B \cdot B M \cos 60^{\circ} \\
=(2 a+b)^{2}+(a+b)^{2}-(2 a+b)(a+b),
\end{array}
$$

which simplifies to $3 a^{2}+3 a b+b^{2}=1$.
Since $C N$ bisects $\angle A C B$, we have
$$
\frac{A N}{N B}=\frac{A C}{B C} \Rightarrow \frac{2 a}{b}=\frac{2}{B C} \text {. }
$$

Thus, $B C=\frac{b}{a}$.
By Menelaus' theorem, we have
$$
\begin{array}{l}
\frac{A B}{B N} \cdot \frac{N X}{X C} \cdot \frac{C M}{M A}=1 \\
\Rightarrow \frac{2 a+b}{b} \cdot \frac{b}{X C}=1 \\
\Rightarrow X C=2 a+b . \\
\text { Also, } \triangle N B C \backsim \triangle X M C \Rightarrow \frac{B C}{M C}=\frac{N C}{X C} \\
\Rightarrow \frac{\frac{b}{a}}{1}=\frac{2 a+2 b}{2 a+b} \Rightarrow b^{2}+2 a b=2 a^{2}+2 a b \\
\Rightarrow b^{2}=2 a^{2} \Rightarrow a=\frac{b}{\sqrt{2}} .
\end{array}
$$

Substituting equation (2) into equation (1), we get
$$
b^{2}=\frac{2}{5+3 \sqrt{2}}=\frac{10-6 \sqrt{2}}{7} .
$$