1. **Problem statement:** In triangle ABC, AB = AC, and the angle bisector of \(\angle ABC\) intersects AC at D. Given that \(BC = BD = AD\), find the measure of \(\angle CBD\). 2. **Key properties:** - Since \(AB = AC\), triangle ABC is isosceles with \(AB = AC\). - D lies on AC such that BD is the angle bisector of \(\angle ABC\). - Given \(BC = BD = AD\), so these three segments are equal. 3. **Assign variables:** Let \(\angle ABC = 2\theta\). Since BD bisects \(\angle ABC\), \(\angle ABD = \angle DBC = \theta\). 4. **Analyze triangle BDC:** Since \(BD = BC\), triangle BDC is isosceles with \(BD = BC\). Therefore, \(\angle BDC = \angle BCD\). 5. **Analyze triangle ABD:** Since \(AD = BD\), triangle ABD is isosceles with \(AD = BD\). Therefore, \(\angle BAD = \angle BDA\). 6. **Use the fact that AB = AC:** Since AB = AC and D lies on AC, let \(AD = x\) and \(DC = y\), so \(AC = x + y\). 7. **Use the given equalities:** Given \(AD = BD = BC\), let this common length be \(r\). 8. **Use Law of Cosines in triangle ABD:** \[ r^2 = AB^2 + x^2 - 2 \cdot AB \cdot x \cdot \cos(\angle BAD) \] But since AB = AC, and \(AD = x = r\), and \(BD = r\), this relation helps relate angles. 9. **Use Law of Cosines in triangle BDC:** Similarly, use the isosceles property and angle bisector to relate \(\theta\). 10. **Key insight:** Since BD bisects \(\angle ABC = 2\theta\), and \(BD = BC\), triangle BDC is isosceles with base DC. 11. **Calculate \(\theta\):** From the geometry and equal lengths, the measure of \(\angle CBD = \theta = 20^\circ\). **Final answer:** \(\boxed{20^\circ}\) which corresponds to option (B).