prove parallelogram.

AD=BC(given)
∠ADB=∠DBC(givven)
BD=BD(given)
∴ΔADB is corngent to ΔCDB (S.A.S.)
∵ΔADB is corngent to ΔCDB
∴∠BCD=∠DAB
∴∠DBA=∠CDB(∠sum ofΔ)
∴∠DBA+∠CBD=∠CDB+∠ADB
∵∠BAD=∠DCB,∠ADC=∠CBA
∴AB//CD, AD//BC
∴ABCD is a parallelogram(definition of //gram)

Yes!you're correct!