Find value of $\frac{\sum AA_1 \cos\left(\frac{A}{2}\right)}{\sum \sin A}$

Triangle $$\Delta ABC$$ is inscribed in a circle of radius one unit. If the internal angle bisectors of angles $$\angle A, \angle B,\angle C$$ meets the circle at the points $$A_1,B_1,C_1$$ respectively. Find value of $$S=\frac{\sum AA_1 \cos\left(\frac{A}{2}\right)}{\sum \sin A}$$

My try:

Letting $$BC=a, AB=c, AC=b$$

We have $$AD=\frac{2bc}{b+c}\cos\left(\frac{A}{2}\right)$$

Hence $$AA_1=AD+DA_1=\frac{2bc}{b+c}\cos\left(\frac{A}{2}\right)+DA_1$$

But how to find $$DA_1$$?

Note that $$A_1,$$ $$B_1,$$ $$C_1$$ are the midpoints of $$\overset{\huge\frown}{BC},$$ $$\overset{\huge\frown}{CA},$$ $$\overset{\huge\frown}{AB}$$ respectively.

Therefore, considering $$\triangle A_1BC,$$ $$A_1B=A_1C$$. Now we can use law of cosines in $$\triangle AA_1B$$ and $$\triangle AA_1C$$ to get,

$$(A_1B)^2=(AA_1)^2+c^2-2\cdot AA_1\cdot c\cos\frac A2$$ $$(A_1C)^2=(AA_1)^2+b^2-2\cdot AA_1\cdot b\cos\frac A2$$

Using these two, we can easily get $$AA_1\cos\left(\frac A2\right)=\frac{(b+c)}2$$

Also from law of sines we have $$\frac{\sin A}a=\frac{\sin B}b=\frac{\sin C}c=2R$$ where $$R$$ is the radius of the circumcircle.

Hence, $$\frac{\sum AA_1\cos\left(\frac A2\right)}{\sum\sin A}\equiv R\cdot\frac{\sum(b+c)}{\sum a}$$

Can you take it from here?