Question

In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that $\angle AOB+\angle BOC+\angle COD+\angle DOE+\angle EOA={360}^{\circ }$.

Answer 1

Produce AO to F such that AOF is a straight line

Now $\angle AOB+\angle BOF={180}^{\circ }$ (Linear pair)
$\Rightarrow \angle AOB+\angle BOC+\angle COF={180}^{\circ }$ ...(i)
Similarly, $\angle AOE+\angle EOF={180}^{\circ }$
$\Rightarrow \angle AOE+\angle EOD+\angle DOF={180}^{\circ }$ ...(ii)
Adding (i) and (ii)
$\angle AOB+\angle BOC+\angle COD+\angle DOE+\angle EOA={360}^{\circ }$
Hence proved.