Question

In the given below fig, rays OA, OB, OC, OP and 0E have the common end point O. Show
that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

Answer 1

Given that

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

A ray of opposite to OA is drawn

Since `∠`AOB, `∠`BOF are linear pairs

`∠`AOB + `∠`BOF = 180°

`∠`AOB + `∠`BOC + `∠`COF = 180°

Also

`∠`AOE, `∠`EOF are linear pairs

`∠`AOE + `∠`EOF = 180°

`∠`AOE + `∠`DOF + `∠`DOE = 180°

By adding (1) and (2) quations we get                       

`∠`AOB + `∠`BOC + `∠`COF + `∠`AOE + `∠`DOF + `∠`DOE = 360°

`∠`AOB + `∠`BOC + `∠`COD + `∠`DOE + `∠`EOA = 360°

Hence proved.

Answer 2

Let us draw AOXa straight line.

∠AOE,∠DOE and ∠DOXform a linear pair. Thus, their sum should be equal to180°.

Or, we can say that:

 ∠AOE +∠DOE +∠DOX  = 180°   (I)

Similarly,, ∠AOB,∠BOC and ∠COXform a linear pair. Thus, their sum should be equal to180°.

Or, we can say that:

 ∠AOB +∠BOC+ ∠COX =  180°       (II)

On adding (I) and (II), we get:

∠AOB +∠BOC + ∠COX +∠DOX +∠AOE +∠DOE = 180°+180°

∠AOB +∠BOC + ∠COD +∠AOE +∠DOE = 360°

Hence proved.