#### Answer

Using method ASA with 2 congruent pairs of angles and a congruent pair of included side, it is proved that $\triangle ABC\cong\triangle DBC$.

#### Work Step by Step

Since $\angle 1$ and $\angle 2$ are both right angles, it is deduced that $\angle 1\cong\angle 2$.
It is also mentioned that
- $\overline{AB}\cong\overline{BD}$
- $\angle A\cong\angle D$
We now have 2 angles and the included side of $\triangle ABC$ are congruent with 2 angles and the included side of $\triangle DBC$. Therefore, according to method ASA, $\triangle ABC\cong\triangle DBC$.