Question

given AB II DE, AB = DE,AC II DF,AC =DE prove that BC II EF BC = EF

Answer 1

Let, Point P be the intersection of AB and DE.

=> ∠APD = 110 degrees Vertically Opposite angles

=> ∠DMB = ∠AME = 70 Degrees

Let, Q be the intersection of AB and DF

In Triangle DPQ,

∠DQP = 180 Degrees - 60 Degrees - 70 Degrees

∠DQP = 50 degrees

=> DFE = 50 Degrees (Interior angles as AB parallel to EF

Similarly you can compute remaining angles.

You will get

x = 120 degrees and y = 120 degrees