#### Answer

Using method SAS for two congruent pairs of sides and one congruent pair of included angle, we can prove that $\triangle ABC\cong\triangle DBC$.

#### Work Step by Step

Since B is the midpoint of $\overline{AD}$, the distance from B to A is equal with the distance from B to D.
In other words, $\overline{BA}\cong\overline{BD}$.
It is also given that
- $\angle A\cong\angle D$
- $\overline{AB}\cong\overline{BD}$
Therefore, we have 2 sides and the included angle of $\triangle ABC$ are congruent with 2 sides and the included angle of $\triangle DBC$.
So, according to method SAS, $\triangle ABC\cong\triangle DBC$.