We are given that $\overline{CA}$ bisects $\angle BAD$ and $\angle BCD$. We want to prove that $\triangle ABC \cong \triangle ADC$.

1. Problem Description

We are given that $\overline{CA}$ bisects $\angle BAD$ and $\angle BCD$. We want to prove that $\triangle ABC \cong \triangle ADC$.

2. Solution Steps

Step 1: $\overline{CA}$ bisects $\angle BAD$ and $\overline{CA}$ bisects $\angle BCD$. Reason: Given.

Step 2: $\angle BAC \cong \angle DAC$ because $\overline{CA}$ bisects $\angle BAD$. Reason: Definition of angle bisector.

$\angle BCA \cong \angle DCA$ because $\overline{CA}$ bisects $\angle BCD$. Reason: Definition of angle bisector.

Step 3: $\overline{AC} \cong \overline{AC}$. Reason: Reflexive Property.

Step 4: $\triangle ABC \cong \triangle ADC$ by Angle-Side-Angle (ASA) congruence. Since $\angle BAC \cong \angle DAC$, $\overline{AC} \cong \overline{AC}$, and $\angle BCA \cong \angle DCA$, we can conclude that $\triangle ABC \cong \triangle ADC$.

3. Final Answer

$\triangle ABC \cong \triangle ADC$