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Answer:
[tex]\sf{ \bar{BC}} = \sf{ \bar{EC}}[/tex]
Explaination:
[tex]\sf \angle EBC = 40 \degree \: ..(given) \\\sf \angle EBC + \angle \: CEB + \angle BCE = 180 \degree \\ \sf 40 + \angle \: CEB + 100 = 180 \: \\ \sf ..(sum \: of \: measure \: of \: all \: angles \: of \: triangle \\ \sf \angle \: CEB + 140 = 180 \\\angle \: \sf \: CEB = 180 - 140 \\ \sf \: \angle \: CEB = 40 \degree \\ \\ \sf \therefore \: \angle EBC = \angle \: CEB \\ \sf \: In \:triangle \: sides \: opposite \: to \: congruent \: angles \: are \: congruent \\ \therefore \: \sf{ \bar{BC}} = \sf{ \bar{EC}}[/tex]
[tex] \sf \pink{hope \: it \: helps :)}[/tex]