since 0<a1<a2<π/2, w1 and w2 are in Q1. furthermore, since they both have the same modulus of 1, you can form a rhombus with diagonals w1+w2 and w1-w2
arg(w1-w2)=arg(w1+w2)-π/2 {rhombus diagonals are perpendicular}
w1+w2 (D) bisects the angle W2OW1 so arg(w1+w2)=1/2(a2-a1)+a1=1/2(a1+a2)...