We are given that $\overline{DC}$ bisects $\angle ACB$ and $\overline{AC} \cong \overline{BC}$. We want to prove that $\triangle ACD \cong \triangle BCD$.

1. Problem Description

We are given that $\overline{DC}$ bisects $\angle ACB$ and $\overline{AC} \cong \overline{BC}$. We want to prove that $\triangle ACD \cong \triangle BCD$.

2. Solution Steps

Step 1 is given:

$\overline{DC}$ bisects $\angle ACB$

$\overline{AC} \cong \overline{BC}$

Reason: Given

Step 2:

$\angle ACD \cong \angle BCD$

Reason: Definition of angle bisector

Step 3:

$\overline{CD} \cong \overline{CD}$

Reason: Reflexive Property of Congruence

Step 4:

$\triangle ACD \cong \triangle BCD$

Reason: Side-Angle-Side (SAS) Congruence Postulate.

3. Final Answer

$\triangle ACD \cong \triangle BCD$