1. **Problem statement:** In circle ABC passing through center O of circle ADC, with ADB a straight line and given \(\angle BDC = 46^\circ\), find \(\angle CBD\). 2. **Key facts and formulas:** - Since ABC passes through O, O lies on circle ABC. - ADB is a straight line, so \(\angle ADB = 180^\circ\). - The angle subtended by the same chord at the circumference is equal. - The angle at the center is twice the angle at the circumference subtended by the same chord. 3. **Step-by-step solution:** - Given \(\angle BDC = 46^\circ\). - Since ADB is a straight line, \(\angle BDA = 180^\circ - 46^\circ = 134^\circ\). - In circle ABC, \(\angle BDA\) and \(\angle BCA\) subtend the same chord BA, so \(\angle BCA = 134^\circ\). - Triangle BCD has angles \(\angle BDC = 46^\circ\), \(\angle BCD = 134^\circ\), and \(\angle CBD = x\). - Sum of angles in triangle BCD: \(x + 46 + 134 = 180\). - Simplify: \(x + 180 = 180\) so \(x = 0^\circ\). This suggests a reconsideration: since \(\angle BCA = 134^\circ\) is large, and B, C, D are points on the circle, the angle \(\angle CBD\) is the angle at B in triangle BCD. Alternatively, use the fact that \(\angle BDC = 46^\circ\) and \(\angle BDA = 134^\circ\) with ADB straight. Since ABC passes through O, and O is center of ADC, chord BC subtends \(\angle BOC\) at center O. Using the property that the angle at center is twice the angle at circumference: \[ \angle BOC = 2 \times \angle BDC = 2 \times 46^\circ = 92^\circ. \] In triangle BOC, \(\angle BOC = 92^\circ\), and OB = OC (radii), so triangle BOC is isosceles. Therefore, \(\angle OBC = \angle OCB = y\). Sum of angles in triangle BOC: \[ 92^\circ + 2y = 180^\circ \implies 2y = 88^\circ \implies y = 44^\circ. \] Since \(\angle CBD = \angle OBC = 44^\circ\), the answer is: \[ \boxed{44^\circ} \].