(8) Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and a regular triangle $M F_{1} F_{2}$ is constructed with the line segment $F_{1} F_{2}$ as one side. If the midpoint of side $M F_{1}$ lies on the hyperbola, then the eccentricity of the hyperbola is ( ).
(A) $4+2 \sqrt{3}$
(B) $\sqrt{3}-1$
(C) $\frac{\sqrt{3}+1}{2}$
(D) $\sqrt{3}+1$
8 Let the intersection point of $M F_{1}$ and the hyperbola be $P$, with foci $F_{1}(-c, 0)$ and $F_{2}(c, 0)$. By plane geometry, we know that $F_{2} P \perp F_{1} M$. Since $\left|F_{1} F_{2}\right|=2 c$, we have
$$
\left|P F_{2}\right|=2 c \sin 60^{\circ}=\sqrt{3} c,\left|P F_{1}\right|=c,
$$

Thus,
$$
2 a=\left|P F_{2}\right|-\left|P F_{1}\right|=\sqrt{3} c-c=(\sqrt{3}-1) c,
$$

Therefore,
$$
e=\frac{c}{a}=\sqrt{3}+1 .
$$

Hence, the answer is D.