1. **Problem Statement:** Prove the following about the circle and tangents: 11.1 PTKQ is a cyclic quadrilateral. 11.2 PK bisects angle $\widehat{TKQ}$. 11.3 $\widehat{A} = \widehat{Q}_3$. 11.4 If $\widehat{T}_2 = \widehat{P}_1$ then $\widehat{AQT} = 90^\circ$. --- 2. **Key Theorems and Properties:** - Tangent-secant theorem: Tangent to a circle is perpendicular to the radius at the point of tangency. - Opposite angles of a cyclic quadrilateral sum to $180^\circ$. - Alternate interior angles are equal when a transversal crosses parallel lines. - Angles subtended by the same chord are equal. --- ### 11.1 Prove PTKQ is cyclic: 3. Since PT and PQ are tangents from P to the circle at T and Q, $\angle PTQ = \angle PQT$ (tangents from a point are equal). 4. PK is parallel to QA, so $\angle PKT = \angle QAT$ (alternate interior angles). 5. Angles $\angle PTQ$ and $\angle PKQ$ subtend the same arc $\widehat{TQ}$. 6. Therefore, $\angle PTQ + \angle PKQ = 180^\circ$ (sum of opposite angles in quadrilateral PTKQ). 7. Hence, PTKQ is cyclic by the cyclic quadrilateral criterion. --- ### 11.2 Prove PK bisects $\widehat{TKQ}$: 8. Since PK is parallel to QA, $\angle PKQ = \angle QAT$ (alternate interior angles). 9. $\angle TKQ$ is split into two equal angles by PK because PK is parallel and intersects chord TQ at R. 10. Therefore, PK bisects $\widehat{TKQ}$. --- ### 11.3 Prove $\widehat{A} = \widehat{Q}_3$: 11. Angles subtended by the same chord are equal. 12. $\widehat{A}$ and $\widehat{Q}_3$ subtend the same chord. 13. Therefore, $\widehat{A} = \widehat{Q}_3$. --- ### 11.4 If $\widehat{T}_2 = \widehat{P}_1$ then $\widehat{AQT} = 90^\circ$: 14. Given $\widehat{T}_2 = \widehat{P}_1$, and since PT and PQ are tangents, angles at T and P relate to the radius. 15. Using the tangent-radius perpendicularity, $\widehat{AQT}$ is a right angle. 16. Hence, $\widehat{AQT} = 90^\circ$. --- **Final answers:** - PTKQ is cyclic. - PK bisects $\widehat{TKQ}$. - $\widehat{A} = \widehat{Q}_3$. - If $\widehat{T}_2 = \widehat{P}_1$ then $\widehat{AQT} = 90^\circ$.