Question

In the figure BA||ED and BC||EF show that angle ABC= angleDEF​

Answer 1

Step-by-step explanation:

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Answer 2

Answer

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