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Question -

In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.



Answer -

Produce AO to F such that AOF is a straight line
Now ∠AOB + ∠BOF = 180° (Linear pair)
 
⇒ ∠AOB + ∠BOC + ∠COF = 180° …(i)
Similarly, ∠AOE + ∠EOF = 180°
⇒ ∠AOE + ∠EOD + ∠DOF = 180° …(ii)
Adding (i) and (ii)
∠AOB + ∠BOC + ∠COF + ∠DOF + ∠EOD + ∠AOE = 180° + 180°
⇒ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
Hence proved.

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