Question

In Fig. BA parallel ED and BC parallel EF. Prove that angle ABC = angle DEF. ​

Answer 1

Hope it is helpful to you

Answer 2

DE intersect BC at P

EF||BC & DP is transversal

∠ DEF = ∠ DPC . . . . . . . . ( i )

Now, AB||DP & BC is transversal

∠ DPC = ∠ ABC . . . . . . . ( ii )

From ( i ) & ( ii ),

we get :

∠ ABC = ∠ DEF

Hence , proved