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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 $$S=\frac{\sum AA_1 \cos\left(\frac{A}{2}\right)}{\sum \sin A}$$ enter image description here

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 $DA_1$?

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