Question

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

Answer 1

Given: Rays $OA,OB,OC,OD\text{and}OE$ have the common endpoint $O.$

Construction: Extend the ray $OA$ to point $X$ which make a straight-line $AX.$


From the figure:

$\angle AOB$ and $\angle BOX$ are forming linear pair,

$\therefore \angle AOB+\angle BOX={180}^{\circ }$

$\Rightarrow \angle AOB+\angle BOC+\angle COX={180}^{\circ }$ $\dots \left(1\right)$

Also,

$\angle AOE$ and $\angle EOX$ are forming linear pair,

$\therefore \angle AOE+\angle EOX={180}^{\circ }$

$\Rightarrow \angle AOE+\angle DOE+\angle DOX={180}^{\circ }$ $\dots \left(2\right)$

By adding equations $\left(1\right)$ and $\left(2\right),$ we get

$\angle AOB+\angle BOC+\angle COX+\angle AOE+\angle DOE+\angle DOX={180}^{\circ }+{180}^{\circ }$

Since, $\angle BOC+\angle DOX=\angle COD$

$\therefore \angle AOB+\angle BOC+\angle COD+\angle DOE+\angle EOA={360}^{\circ }$

Hence, proved.