1. **Problem statement:** Given a circle with center O, quadrilateral ABCD inscribed in the circle, and a tangent line PBQ touching the circle at B, we need to prove: (i) $\angle ADC = \angle ABP + \angle CBQ$ (ii) If $AB = CB$, then $\angle AÔC = 4 \times \angle CBQ$ 2. **Key concepts and formulas:** - The angle between a tangent and a chord through the point of contact equals the angle subtended by the chord in the alternate segment (Alternate Segment Theorem). - Opposite angles of a cyclic quadrilateral sum to $180^\circ$. 3. **Proof of (i):** - By the Alternate Segment Theorem, $\angle ABP$ is equal to the angle subtended by chord $AB$ in the alternate segment, which is $\angle ADB$. - Similarly, $\angle CBQ$ equals the angle subtended by chord $CB$ in the alternate segment, which is $\angle BDC$. - Since $\angle ADC = \angle ADB + \angle BDC$ (angles around point D on the circle), we have: $$\angle ADC = \angle ABP + \angle CBQ$$ 4. **Proof of (ii) when $AB = CB$: ** - Since $AB = CB$, triangle $ABP$ and $CBQ$ are symmetric with respect to line $OB$. - From (i), $\angle ADC = \angle ABP + \angle CBQ = 2 \times \angle CBQ$ (because $\angle ABP = \angle CBQ$). - In cyclic quadrilateral $ABCD$, $\angle AÔC$ is the central angle subtending arc $AC$. - The inscribed angle $\angle ADC$ subtends the same arc $AC$, so: $$\angle AÔC = 2 \times \angle ADC = 2 \times (2 \times \angle CBQ) = 4 \times \angle CBQ$$ **Final answers:** (i) $\angle ADC = \angle ABP + \angle CBQ$ (ii) If $AB = CB$, then $\angle AÔC = 4 \times \angle CBQ$