#### Answer

Using method ASA for 2 congruent pairs of angles and one included pair of sides, we can prove that $\triangle ABC\cong\triangle DBC$.

#### Work Step by Step

Since $\vec{CB}$ bisects $\angle ACD$, two resulting angles of this bisect would be equal with each other.
That means $\angle 3\cong\angle 4$.
It is also mentioned that
- $\angle A\cong\angle D$
- $\overline{AC}\cong\overline{CD}$
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, $\triangle ABC\cong\triangle DBC$.