Reflect ΔECD over segment AC. This establishes that . Then, translate point E' to point A. This establishes that . Therefore, ΔACB ~ ΔECD by the AA similarity postulate.
a. ∠ABC ≅ ∠E'D'C'; ∠E'D'C' ≅ ∠ABC
b. ∠ACB ≅ ∠E'C'D'; ∠D'E'C' ≅ ∠BAC
c. ∠ACB ≅ ∠E'C'D'; ∠E'D'C' ≅ ∠ABC
d. ∠ABC ≅ ∠E'D'C'; ∠D'E'C' ≅ ∠BAC