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

Given:

Rays $OA, OB, OC, OD$ and $OE$ have the common end point $O$.

To do:

We have to show that $\angle AOB + \angle BOC + \angle COD + \angle DOE + \angle EOA = 360^o$.

Solution:

Produce $AO$ to $F$ such that $AOF$ is a straight line.

This implies,

$\angle AOB + \angle BOF = 180^o$               (Linear pair)

$\angle AOB + \angle BOC + \angle COF = 180^o$....…(i)

Similarly,

$\angle AOE + \angle EOF = 180^o$

$\angle AOE + \angle EOD + \angle DOF = 180^o$....…(ii)

Adding (i) and (ii)

$\angle AOB + \angle BOC + \angle COF + \angle DOF + \angle EOD + \angle AOE = 180^o + 180^o$

$\angle AOB + \angle BOC + \angle COD + \angle DOE + \angle EOA = 360^o$

Hence proved.

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Updated on: 10-Oct-2022

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