Answer :

Given: AB and OF are two lines cut by the transversal BO. CE is joined. ∠ABO = 65° and ∠FEC = 145°.

Construction: Produce FE to cut BC at O.

Proof: In Δ COE, ∠FEC = 145° is the exterior angle.

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145° = ∠EOC + 30°

∠EOC = 145° - 30° = 115° --------------- (1)

∠BOE and ∠EOC form a linear pair.

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∠BOE = 180° - 115°

= 65°

AB and OF are two lines cut by the transversal BO.

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and ∠ BOE = ∠ ABO = 65°

Since alternate interior angles are equal.

.

Construction: Produce FE to cut BC at O.

Proof: In Δ COE, ∠FEC = 145° is the exterior angle.

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^{.}. ∠FEC = ∠EOC + ∠OCE (In a Δ the exterior angle is equal to sum of interior opposite angles).145° = ∠EOC + 30°

∠EOC = 145° - 30° = 115° --------------- (1)

∠BOE and ∠EOC form a linear pair.

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^{.}. ∠BOE + ∠EOC = 180° (Linear Pair Axiom).∠BOE = 180° - 115°

= 65°

AB and OF are two lines cut by the transversal BO.

.

^{.}. ∠ ABO and ∠ BOE are alternate angle,and ∠ BOE = ∠ ABO = 65°

Since alternate interior angles are equal.

.

^{.}. AB || EF.Rate this question :

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