Question 1170429
Here's the step-by-step proof:

**Given:**

1.  $\overline{CA}$ bisects $\angle BAD$.
2.  $\angle B \cong \angle D$.

**Prove:**

$\triangle ABC \cong \triangle ADC$.

**Proof:**

| **Statement** | **Reason** |
|---|---|
| 1. $\overline{CA}$ bisects $\angle BAD$. | Given |
| 2. $\angle BAC \cong \angle DAC$. | Definition of angle bisector. |
| 3. $\angle B \cong \angle D$. | Given |
| 4. $\overline{CA} \cong \overline{CA}$. | Reflexive Property of Congruence. |
| 5. $\triangle ABC \cong \triangle ADC$. | Angle-Side-Angle (ASA) Congruence Theorem. |

**Explanation:**

1.  We are given that $\overline{CA}$ bisects $\angle BAD$. This means that $\overline{CA}$ divides $\angle BAD$ into two congruent angles.
2.  By the definition of an angle bisector, $\angle BAC$ and $\angle DAC$ are congruent.
3.  We are given that $\angle B \cong \angle D$.
4.  $\overline{CA}$ is a shared side between $\triangle ABC$ and $\triangle ADC$. By the Reflexive Property of Congruence, any segment is congruent to itself.
5.  We have shown that two angles ($\angle BAC$ and $\angle B$) and the included side ($\overline{CA}$) of $\triangle ABC$ are congruent to the corresponding parts ($\angle DAC$, $\angle D$, and $\overline{CA}$) of $\triangle ADC$. Therefore, by the Angle-Side-Angle (ASA) Congruence Theorem, $\triangle ABC \cong \triangle ADC$.