AC = BC, angle ABD = 20 degrees. Find angle CBE.

Solution:

1. The problem states that AC = BC, which means that triangle ABC is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, \(\angle BAC = \angle ABC\).
2. We are given that \(\angle ABD = 20^{\circ}\).
3. We need to find \(\angle CBE\). Note that \(\angle ABC\) and \(\angle CBE\) form a linear pair, meaning they are supplementary and their sum is \(180^{\circ}\). However, the diagram suggests that E is a point on the extension of AB, making \(\angle CBE\) an exterior angle to \(\triangle ABC\) if C were to the right of B, or an angle adjacent to \(\angle ABC\) if E is on the extension of AB. Based on the arc symbol at B and the position of E, it's likely that \(\angle CBE\) refers to the angle supplementary to \(\angle ABC\) on the line AE. Let's assume E is on the extension of the line segment AB. So \(\angle ABC + \angle CBE = 180^{\circ}\).
4. Let \(\angle BAC = \angle ABC = x\).
5. In \(\triangle ABC\), the sum of angles is \(180^{\circ}\). So, \(\angle BAC + \angle ABC + \angle ACB = 180^{\circ}\).
6. Substituting \(\angle BAC = x\) and \(\angle ABC = x\), we get \(x + x + \angle ACB = 180^{\circ}\), which simplifies to \(2x + \angle ACB = 180^{\circ}\).
7. From the diagram, it appears that D is a point on AC and BD is perpendicular to AC (indicated by the right angle symbol). This means \(\angle BDA = 90^{\circ}\).
8. In \(\triangle ABD\), the sum of angles is \(180^{\circ}\). So, \(\angle BAD + \angle ABD + \angle BDA = 180^{\circ}\).
9. We are given \(\angle ABD = 20^{\circ}\) and \(\angle BDA = 90^{\circ}\). Also, \(\angle BAD = \angle BAC = x\).
10. Substituting these values into the equation for \(\triangle ABD\): \(x + 20^{\circ} + 90^{\circ} = 180^{\circ}\).
11. Solving for x: \(x + 110^{\circ} = 180^{\circ}\) => \(x = 180^{\circ} - 110^{\circ}\) => \(x = 70^{\circ}\).
12. So, \(\angle BAC = 70^{\circ}\) and \(\angle ABC = 70^{\circ}\).
13. Now we need to find \(\angle CBE\). Since E is on the extension of the line segment AB, \(\angle ABC\) and \(\angle CBE\) are supplementary angles.
14. Therefore, \(\angle CBE = 180^{\circ} - \angle ABC\).
15. \(\angle CBE = 180^{\circ} - 70^{\circ}\).
16. \(\angle CBE = 110^{\circ}\).

Answer: 110