The rays \(OA\), \(OB\), \(OC\), and \(OD\) originate from \(O\). Prove that \(\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^\circ\).
Proof:
We should prove that \(\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^\circ\).
Let us extend \(OC\) to \(E\) as given in the following figure.
Here, \(EOC\) is a straight line.
\(\angle EOA + \angle AOB + \angle BOC =\) \(^\circ \longrightarrow (1)\) [By linear pair of angles axiom]
\(\angle EOD + \angle COD =\) \(^\circ \longrightarrow (2)\) [By linear pair of angles axiom]
On adding \((1)\) and \((2)\), we get:
\(\angle EOA + \angle AOB + \angle BOC + \angle EOD + \angle COD =\)\(^\circ +\)\(^\circ\)
\(\angle EOA + \angle AOB + \angle BOC + \angle EOD + \angle COD = \)\(^\circ \longrightarrow (3)\)
\(\angle DOA = \angle EOD + \angle EOA\)
On substituting \(\angle DOA = \angle EOD + \angle EOA\) in \((3)\), we get:
\(\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^\circ\)
Hence, the required condition is proved.