Triangle given the angle A, the A-bisector d, and the sum b+c of the sides

1) Place the angle A and the bisector AD. On a side of the angle take M such that AM = (b+c)/2.
2) The perpendicular from M to AM cuts the bisector AD at P.
3) The arc that inscribes an angle A/2 on the segment DP cuts the side of M at C and B', C the nearest to A.
4) Let B be the reflection of B' on AD. Then ABC is the requested triangle.