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Ray OE bisects ∠AOB and OF is the ray opposite to OE. Show that ∠FOB = ∠FOA.


∠FOB + ∠BOE = 180° ...(1)
| Linear Pair Axiom
∠FOA + ∠AOE = 180° ...(2)
| Linear Pair Axiom
From (1) and (2),
∠FOB + ∠BOE = ∠FOA + ∠AOE ...(3)
But    ∠BOE = ∠AOE
| ∵ Ray OE bisects ∠AOB
∴ From (3),
⇒    ∠FOB = ∠FOA.

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


In figure, if y = 20°, prove that the line AOB is a straight line.

Prove “if two lines intersect each other, then the vertically opposite angles are equal.”

In figure, OP bisects ∠AOC, OQ bisects ∠BOC and OP ⊥ OQ. Show that the points A, O and B are collinear.


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