Given that $\overline{CA}$ bisects $\angle BAD$ and $\overline{CA}$ bisects $\angle BCD$, we want to prove that $\triangle ABC \cong \triangle ADC$.

1. Problem Description

Given that $\overline{CA}$ bisects $\angle BAD$ and $\overline{CA}$ bisects $\angle BCD$, 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: $\angle BAC \cong \angle DAC$ and $\angle BCA \cong \angle DCA$. Reason: Definition of angle bisector.

If a ray bisects an angle, then it divides the angle into two congruent angles.

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

Step 4: $\triangle ABC \cong \triangle ADC$. Reason: Angle-Side-Angle (ASA) congruence. We have $\angle BAC \cong \angle DAC$, $\overline{AC} \cong \overline{AC}$, and $\angle BCA \cong \angle DCA$, so we can apply ASA.

3. Final Answer

$\angle BAC \cong \angle DAC$ and $\angle BCA \cong \angle DCA$