Question

Hii anyone can solve it

Please don't cheat ​

Answer 1

Answer:

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Given :- AOB is a straight line and rays OC , OD and OE stand on it , forming Angle AOC , Angle COD and DOE and And Angle EOB.

To prove :- Angle AOC + Angle COD + Angle EOB + Angle DOE = 180°

Proof :- Ray OC stands on line AB.

Therefore,

Angle AOC + Angle COB = 180°

=> Angle AOC + ( Angle COD + Angle DOE + Angle EOB ) = 180° [ Since Angle COB = Angle COD + Angle DOE + Angle EOB )

=> Angle AOC + Angle COD + Angle DOE + Angle EOB = 180°

Hence,

The sum of all the angles formed on the same side of a line AB at a point O on it is 180°.

Answer 2

$\Large\boxed{SOLUTION }$

Refer to attached Image

Given = AOM is a straight line and ray OP , OQ and OR stand on it

To Prove = ∠AOP + ∠POQ + ∠QOR + ∠ROM = 180

Proof = Ray OP stands on line AM

∠AOP + ∠POM = 180

∠AOP + ∠POQ + ∠QOR + ∠ROM = 180 [Explanation ∠POQ + ∠QOR + ∠ROM = ∠POM]

∠AOP + ∠POQ + ∠QOR + ∠ROM = 180

Therefore

Sum of all angles formed on same side of line AM at a point O on it is 180