1. **State the problem:** Given that $\overline{AB} \cong \overline{CB}$ and $\overline{AD} \cong \overline{CD}$, prove that $\overline{DB}$ bisects $\angle ABC$. 2. **Given:** - $\overline{AB} \cong \overline{CB}$ (Given) - $\overline{AD} \cong \overline{CD}$ (Given) 3. **Goal:** Show that $\overline{DB}$ divides $\angle ABC$ into two equal angles, i.e., $\angle ABD \cong \angle DBC$. 4. **Proof steps:** 1. Consider triangles $\triangle ABD$ and $\triangle CBD$. 2. We know $\overline{AB} \cong \overline{CB}$ (Given). 3. We know $\overline{AD} \cong \overline{CD}$ (Given). 4. $\overline{DB}$ is common to both triangles, so $\overline{DB} \cong \overline{DB}$ (Reflexive property). 5. By the Side-Side-Side (SSS) congruence postulate, $\triangle ABD \cong \triangle CBD$. 6. Corresponding parts of congruent triangles are congruent (CPCTC), so $\angle ABD \cong \angle DBC$. 7. Therefore, $\overline{DB}$ bisects $\angle ABC$. This completes the proof that $\overline{DB}$ bisects $\angle ABC$ using triangle congruence without relying on quadrilateral properties.