Given that,

The rays OA, OB, OC, OD and OE have the common end point O.

A ray of opposite to OA is drawn

Since,

∠AOB, ∠BOF are linear pair

∠AOB + ∠BOF = 180^o

∠AOB + ∠BOC + ∠COF = 180^o (i)

Also,

∠AOE + ∠EOF = 180^o

∠AOE + ∠DOF + ∠DOE = 180^o (ii)

Adding (i) and (ii), we get

∠AOB + ∠BOC + ∠COF + ∠AOE + ∠DOF + ∠DOE = 360^o

∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360^o

Hence, proved