2. Let the three sides of $\triangle A B C$ be $a, b, c$, and the radius of its circumcircle be $R$. If $R=\frac{a \sqrt{b c}}{b+c}$, then the largest angle of $\triangle A B C$ is ( ).
(A) $75^{\circ}$
(B) $120^{\circ}$
(C) $90^{\circ}$
(D) $150^{\circ}$
2.C.

From the given, we know that $(b-c)^{2} \geqslant 0$, which means $b+c \geqslant 2 \sqrt{b c}$.
Therefore, $\frac{\sqrt{b c}}{b+c} \leqslant \frac{1}{2}$.
From the known condition, $\frac{R}{a}=\frac{\sqrt{b c}}{b+c} \leqslant \frac{1}{2}$.
Thus, $a \geqslant 2 R$.
Also, $a \leqslant 2 R$, so $a=2 R$.
Hence, $\frac{2 \sqrt{b c}}{b+c}=1$, which means $(\sqrt{b}-\sqrt{c})^{2}=0$.
Therefore, $b=c$.
Thus, from $a=2 R$ and $b=c$, we can conclude that the given triangle is an isosceles right triangle.