Given; point b.e.c.f is on the agreed line, ab = De, angle a = angle D, AC bisects DF; triangle ABC congruent triangle def, be = CF

Given; point b.e.c.f is on the agreed line, ab = De, angle a = angle D, AC bisects DF; triangle ABC congruent triangle def, be = CF

Parallel proof: ∵ AB De, ∵ B = ∵ def ∵ be = CF, ∵ BC = EF, ≌ ABC ≌ def (SAS) ≌ ACB = ∵ F, ∥ AC DF. Do you need to send the graph to have a look