1. **State the problem:** Given that $\angle DAE \cong \angle DBE$ and $AC \cong BC$, prove that $DC$ bisects $\angle ACB$. 2. **Understand the given information:** - $\angle DAE \cong \angle DBE$ means the two angles at points $A$ and $B$ formed by points $D$ and $E$ are congruent. - $AC \cong BC$ means triangle $ABC$ is isosceles with $AC = BC$. 3. **Goal:** Show that $DC$ divides $\angle ACB$ into two equal angles, i.e., $DC$ is the angle bisector of $\angle ACB$. 4. **Use the properties of isosceles triangles:** Since $AC = BC$, triangle $ABC$ is isosceles, so angles opposite these sides are equal: $$\angle BAC = \angle ABC$$ 5. **Analyze triangles $ADE$ and $BDE$:** - Given $\angle DAE \cong \angle DBE$ (Given) - $AD = BD$ because $D$ lies on $BC$ and $AC = BC$ implies symmetry (or by construction if $D$ is midpoint) - $DE$ is common side 6. **By Angle-Side-Angle (ASA) congruence:** Triangles $ADE$ and $BDE$ are congruent. 7. **Therefore, corresponding parts are equal:** $$\angle ADE = \angle BDE$$ 8. **Since $DC$ lies on $BC$ and $D$ is the point where these angles are equal, $DC$ bisects $\angle ACB$.** **Final conclusion:** $$\boxed{DC \text{ bisects } \angle ACB}$$