Question

In figure ab||ed and bc||ef.show that angle abc=angle def

Answer 1

first , angle afe = angle def - (I) by alt. int. angles.

second , angle abc =angle bcd -(2)

by comparing (1) & (2)

angle abc =angle afe

angle abc =angle def ( afe =def )

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