$23 . \star \star \triangle A B C$ is inscribed in a unit circle, the angle bisectors of the three interior angles $A, B, C$ are extended to intersect this circle at $A_{1}, B_{1}, C_{1}$. Then the value of $\frac{A A_{1} \cdot \cos \frac{A}{2}+B B_{1} \cdot \cos \frac{B}{2}+C C_{1} \cdot \cos \frac{C}{2}}{\sin A+\sin B+\sin C}$ is ).
(A) 2
(B) 4
(C) 6
(D) 8
Parse as shown in the figure, connect $B A_{1}$, then $A A_{1}=2 \sin \left(B+\frac{A}{2}\right)=$ $2 \sin \left(\frac{A+B+C}{2}+\frac{B}{2}-\frac{C}{2}\right)=2 \cos \left(\frac{B}{2}-\frac{C}{2}\right)$. Therefore,
$$
\begin{aligned}
A A_{1} \cdot \cos \frac{A}{2} & =2 \cos \left(\frac{B}{2}-\frac{C}{2}\right) \cos \frac{A}{2}=\cos \frac{A+B-C}{2}+\cos \frac{A+C-B}{2} \\
& =\cos \left(\frac{\pi}{2}-C\right)+\cos \left(\frac{\pi}{2}-B\right)=\sin C+\sin B,
\end{aligned}
$$

Similarly, $B B_{1} \cdot \cos \frac{B}{2}=\sin A+\sin C, C C_{1} \cdot \cos \frac{C}{2}=\sin A+\sin B$,
so $A A_{1} \cdot \cos \frac{A}{2}+B B_{1} \cdot \cos \frac{B}{2}+C C_{1} \cdot \cos \frac{C}{2}=2(\sin A+\sin B+\sin C)$, thus, the original expression $=2$. Therefore, the answer is A.