Angle $C$ of an isosceles triangle $ABC$ equals $120^o$. Each of two rays emitting from vertex $C$ (inwards the triangle) meets $AB$ at some point ($P_i$) reflects according to the rule the angle of incidence equals the angle of reflection" and meets lateral side of triangle $ABC$ at point $Q_i$ ($i = 1,2$). Given that angle between the rays equals $60^o$, prove that area of triangle $P_1CP_2$ equals the sum of areas of triangles $AQ_1P_1$ and $BQ_2P_2$ ($AP_1 < AP_2$).