Question

In the given figure, prove that $\angle AOB+\angle BOC+\angle COD+\angle DOA={360}^{\circ }.$

Answer 1

Adding (1) and (2), we get

∠AOB + ∠BOC + (∠COE + ∠DOE) + ∠AOD = 180° + 180°

⇒ ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (∠COE + ∠DOE = ∠COD)