Extended Bisector

By observing triangle $ABA'$ , we have $\dfrac{AA'}{\sin( B + \frac{1}{2}A)} = 2r =2 \times 1 = 2 \implies AA'=2\sin( B + \frac{1}{2}A)$

The same process for triangle $BCB'$ and $CAC'$ , we have $BB'=2 \sin( C + \frac{1}{2}B)$ and $CC'=2 \sin( A + \frac{1}{2}C)$

Then, $\dfrac{AA'\cos\frac{A}{2}+BB'\cos\frac{B}{2}+CC'\cos\frac{C}{2}}{\sin A+\sin B+\sin C}$ is

$\begin{aligned} & =\dfrac{2\sin( B + \frac{1}{2}A)\cos\frac{A}{2}+2\sin( C + \frac{1}{2}B)\cos\frac{B}{2}+2\sin( A + \frac{1}{2}C)\cos\frac{C}{2}}{\sin A+\sin B+\sin C} \\ & =\dfrac{\sin( B + A)+\sin B+\sin( C + B)+\sin C+\sin( A + C)+sinA}{\sin A+\sin B+\sin C} &\small \color{#3D99F6}{ 2 \sin A \cos B = \sin (A+B) + \sin (A-B)} \\ & =\dfrac{\sin(180-C)+\sin(180-A)+\sin(180-B)+\sin A+\sin B+\sin C}{\sin A+\sin B+\sin C} &\small \color{#3D99F6}{ A+B+C=180 }\\ & =\dfrac{2(\sin A+\sin B+\sin C)}{\sin A+\sin B+\sin C} \\ & =\boxed{2} \end{aligned}$

Observe that $A'B=A'C=\dfrac{a}{2\cos\tfrac{A}{2}}$ . By Ptolemy's Theorem, $AB\cdot A'C+AC\cdot A'B=AA'\cdot BC$ . Substitution and rearrangement yield $AA'\cos\tfrac{A}{2}=\tfrac{b+c}{2}$ and similarly for $BB'$ and $CC'$ . Therefore, the desired sum is $\dfrac{\tfrac{b+c}{2}+\tfrac{c+a}{2}+\tfrac{a+b}{2}}{\sin A+\sin B+\sin C}=\dfrac{a+b+c}{\sin A+\sin B+\sin C}$ . By the Extended Law of Sines, $a=2\sin A$ and similarly for $b$ and $c$ , so the answer is $\boxed{2}$ .

With equilateral triangle, denote R is radius of circumscribed circle. it has: $AA'=BB'=CC'=2R, sin A=sin B=sin C=cos(\frac{A}{2})=cos(\frac{B}{2})=cos(\frac{C}{2})=\frac{\sqrt{3}}{2}$ $\Rightarrow Result=\frac{3\sqrt{3}}{\frac{3\sqrt{3}}{2}}=\boxed{2}$

Well, a short way to do it can be to assume the triangle to be equilateral... The expression then reduces to $2R$ Where $R$ is the radius of the circumcircle , which here is "1" Hence answer = 2