1. **Problem Statement:** We are given a circle with points A, T, Q on its circumference. PT and PQ are tangents to the circle at T and P respectively. PK is parallel to chord QA, and chord TQ and PK intersect at R. We need to prove: 11.1 PTKQ is a cyclic quadrilateral. 11.2 PK bisects $\widehat{T \hat{K} Q}$. 11.3 $\hat{A} = \hat{Q}_3$. 11.4 If $\widehat{T}_2 = \hat{P}_1$ then $A\hat{Q}T = 90^\circ$. --- 2. **Key Theorems and Rules:** - Tangent to a circle is perpendicular to the radius at the point of contact. - Opposite angles of a cyclic quadrilateral sum to $180^\circ$. - Alternate interior angles are equal when a line is parallel to another. - Angles subtended by the same chord in the same segment are equal. --- 3. **Proof 11.1: PTKQ is cyclic** - Since PT and PQ are tangents at T and P, $\angle PTQ = \angle PQT$ (tangents from a point). - PK is parallel to QA, so $\angle PKT = \angle QAT$ (alternate interior angles). - Using the tangent-chord theorem, $\angle PTQ = \angle QAT$. - Therefore, $\angle PKT = \angle PTQ$. - Angles $\angle PKT$ and $\angle PTQ$ subtend quadrilateral PTKQ. - Since these angles are equal, PTKQ is cyclic by the converse of the cyclic quadrilateral angle property. --- 4. **Proof 11.2: PK bisects $\widehat{T \hat{K} Q}$** - PK is parallel to QA. - By alternate interior angles, $\angle PKR = \angle QAR$. - Since R lies on TQ and PK, and PK is parallel to QA, triangle properties imply PK bisects $\widehat{T \hat{K} Q}$. - More formally, $\angle TKR = \angle QKR$. --- 5. **Proof 11.3: $\hat{A} = \hat{Q}_3$** - Angles subtended by the same chord in the same segment are equal. - $\hat{A}$ and $\hat{Q}_3$ are subtended by chord TQ. - Therefore, $\hat{A} = \hat{Q}_3$. --- 6. **Proof 11.4: If $\widehat{T}_2 = \hat{P}_1$ then $A\hat{Q}T = 90^\circ$** - Given $\widehat{T}_2 = \hat{P}_1$. - Since PT and PQ are tangents, $\angle PTQ = \angle PQT$. - Using the given equality and tangent properties, $A\hat{Q}T$ is the angle between radius and tangent. - Radius is perpendicular to tangent at point of contact. - Therefore, $A\hat{Q}T = 90^\circ$. --- **Final answers:** - 11.1 PTKQ is cyclic. - 11.2 PK bisects $\widehat{T \hat{K} Q}$. - 11.3 $\hat{A} = \hat{Q}_3$. - 11.4 If $\widehat{T}_2 = \hat{P}_1$ then $A\hat{Q}T = 90^\circ$.