In figure, show that AB || EF.
          

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.
.^.. ∠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.
    .^.. ∠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.