Given a triangle \( A B C \) and point \( D \) on \( A B \) such that \( C D \) is the angle bisector \( \angle A C B \). Point \( E \) lies on \( A C \) such that \( D E \) is parallel to \( B \) Let \( F \) be on \( B C \) such that \( E F \) is the angle bisector of \( \angle C E D \). Given that the point of intersection between lines \( C D \) and \( E F \) is \( G \). find \( \angle C G E \) in degrees.

To solve for \( \angle C G E \), we start by analyzing the given triangle \( ABC \) and the points defined in the problem. 1. **Understanding the Configuration**: - \( D \) is on \( AB \) such that \( CD \) is the angle bisector of \( \angle ACB \). - \( E \) is on \( AC \) such that \( DE \) is parallel to \( BC \). - \( F \) is on \( BC \) such that \( EF \) is the angle bisector of \( \angle CED \). - The intersection of lines \( CD \) and \( EF \) is point \( G \). 2. **Using Angle Properties**: - Since \( CD \) is the angle bisector of \( \angle ACB \), we have: \[ \angle ACD = \angle BCD \] - Since \( DE \parallel BC \), by the Alternate Interior Angles Theorem, we have: \[ \angle CDE = \angle CBA \] - Let \( \angle ACB = \theta \). Then, we can denote: \[ \angle ACD = \angle BCD = \frac{\theta}{2} \] 3. **Analyzing \( \angle CED \)**: - Since \( DE \parallel BC \), we can also express \( \angle CED \) in terms of \( \theta \): \[ \angle CED = \angle CBA \] - Let \( \angle CBA = \phi \). Thus, we have: \[ \angle CED = \phi \] 4. **Finding \( \angle CEG \)**: - Since \( EF \) is the angle bisector of \( \angle CED \), we have: \[ \angle CEG = \angle CED / 2 = \frac{\phi}{2} \] 5. **Finding \( \angle C G E \)**: - Now, we can find \( \angle CGE \): \[ \angle CGE = \angle CED - \angle CEG = \phi - \frac{\phi}{2} = \frac{\phi}{2} \] 6. **Final Calculation**: - Since \( \angle CEG = \frac{\phi}{2} \) and \( \angle C G E = \frac{\phi}{2} \), we can conclude that: \[ \angle C G E = \frac{\phi}{2} \] 7. **Conclusion**: - The angle \( \angle C G E \) is half of \( \angle CBA \). However, we need to express this in terms of the triangle's angles. Since \( \angle ACB + \angle CBA + \angle CAB = 180^\circ \), we can find that: \[ \angle C G E = 30^\circ \] - Therefore, the final answer is: \[ \boxed{30} \]