As ∠AEC = ∠ADB,
∠CEB = ∠BDC
and since, BE = CD
in ΔBDC and ΔCEB that shares a side of BC are congruent triangles.
Hence, ∠DBC = ∠ECB

In ΔADB and ΔAEC that shares the ∠DAE
as ∠AEC = ∠ADB,
∠DBA = ∠ECA

Hence, ∠ABC = ∠ACB
Hence, ΔABC is isosceles.