BP = EP, where P is the point of intersection of BC and ED produced.
On △ABC and △AED,
AB=AE
∠ABC=∠AED
and BC=ED
∴△ABC≅△AED (by SAS congruence rule.)
AC=AD⟶(1) (By CPCT.)
and ∠BCA=∠EDA⟶(2) (By CPCT.)
In △CAD,AC=AD⟶(from(1))
So, from base angle theorem,
∠ACD=∠ADC⟶(3)
∠BCD=∠BCA+∠ACD⟶(4)
and ∠EDC=angleEDA+∠ADC∴∠EDC=∠BCA+∠ACD⟶(5)⟹∠BCD=∠EDC