A massless inextensible string is wrapped around a frictionless circular peg. The string is taut, with tension $T_2$ and $T_1$ at the points where it leaves of the pg as shown. The segment wrapped around the peg exerts a continuum of contact forces on the peg, perpendicular to the tangent to the surface of the peg at each point. As a result the peg exerts a net reaction force $R$ on the segment which is the sum of this continuum of contact forces (with directions reversed).
The section of string wrapped around the peg is in mechanical equilibrium. WITHOUT assuming $T_1 = T_2$ (I am attempting to deduce that based on the direction of $R$), how can I deduce that $R$ must be perpendicular to the surface of the peg where the centre of mass of the segment wrapped around it is?
Alternatively, can one prove that $T_1 = T_2$ without assuming that $R$ is perpendicular? Arguments involving torque on the peg do not work in this case as the peg is frictionless.