The bisectors of the angles $$A, B, C$$ of the triangle $$ABC$$ meet the circum circle of the triangle again at the points $$D, E, F$$ respectively. What is the value of $$\frac{AD \cos \frac{A}{2} + BE \cos \frac{B}{2} + CF \cos \frac{C}{2}}{\sin A + \sin B + \sin C}$$ if the circum radius of $$ABC$$ is 1 ?

Since $$\angle ABD = \angle ABC + \angle CBD = B + \frac{A}{2}$$, the chord $$AD = 2R \sin\left(B + \frac{A}{2}\right) = 2 \sin\left(B + \frac{A}{2}\right)$$ when $$R = 1$$. Then $$AD \cos \frac{A}{2} = 2 \sin\left(B+\frac{A}{2}\right)\cos\frac{A}{2} = \sin(A+B) + \sin B = \sin C + \sin B$$, and similarly for the other two terms. Adding gives $$2(\sin A + \sin B + \sin C)$$, so the required value is 2.