## Elementary Geometry for College Students (7th Edition) Clone

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