Find the value of x in the following figures if
(a) AB || CD
(b) AB || CD and BC || DE
(c) AB || CD and BC || ED
(d) BA || CD
   
(a)                              (b)

(c)                          (d)

(a) Draw EF || AB

Since AB || EF and AE is a transversal
     .^.. ∠A + ∠1 = 180° 
⇒ 120° + ∠1 = 180° Since AB || CD and AB || EF 
∠1 = 180 - 120 = 60° ….(i)
Since AB || CD and AB || EF 
.^.. CD || EF. Also EC is a transversal
      ⇒ ∠2 + ∠C = 180° ….(Consecutive interior angles)
   ∠2 + 125° = 180°
.^.. ∠2 = 180 - 125 = 55° ….(ii)
Adding equations (i) and (ii), we get .^.. ∠1 + ∠2 = 60° + 55° = 115°
           .^.. x  =  ∠1 + ∠2 = 115°
(b)
Since AB || CD and BC is a transversal 
    .^.. ∠B = ∠C = 40° ….(Alternate angles)
Now BC || DE and CD is transversal 
⇒ ∠C + ∠CDE = 180° ….(Consecutive interior angles)
⇒ 40° + ∠CDE = 180°     .^.. ∠CDE = x = 180° - 40° = 140°
(c)
Since AB || CD and BC is the transversal     ⇒ ∠B = ∠C .....(Alternate angles)
but ∠B = 75°,
.^..  ∠C = 75°
Again we have BC || ED with transversal CD.
 .^..  ∠C + ∠D = 180° …. (Consecutive interior angles) ⇒ 75° + ∠D = 180°      .^.. ∠D = x = 180° - 75° = 105°
(d)
Through O draw OE parallel to BA || CD 
         ⇒ ∠1 + ∠B = 180°          ⇒ ∠1 + 55° = 180° .^.. ∠1 = 180° - 55° = 125° ….(i)
    Again ∠2 + ∠C = 180° ….(Consecutive interior angles)           .^.. ∠2 + 50° = 180°    .^.. ∠2 = 180 - 50 = 130° ….(ii)
Adding (i) and (ii), we get
.^.. ∠1 + ∠2 = x = 125° + 130° = 255°
Hence x = 255°