Prove that : $$ 10.1 \quad \mathrm{PTKQ} $$ is a cyclic quadrilateral 10.2 $$ \begin{array}{ll}\mathrm{PK} ext { bisects } T K \hat{Q} . \ 10.3 & \hat{\mathrm{~A}}=\hat{\mathrm{Q}}_{3} . \ 10.4 & ext { If } \hat{\mathrm{T}}_{2}=\hat{\mathrm{P}}_{1} ext { then } \mathrm{A} \hat{\mathrm{Q} T}=90^{\circ} .\end{array} $$

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Prove that : $$ 10.1 \quad \mathrm{PTKQ} $$ is a cyclic quadrilateral 10.2 $$ \begin{array}{ll}\mathrm{PK} 	ext { bisects } T K \hat{Q} . \ 10.3 & \hat{\mathrm{~A}}=\hat{\mathrm{Q}}_{3} . \ 10.4 & 	ext { If } \hat{\mathrm{T}}_{2}=\hat{\mathrm{P}}_{1} 	ext { then } \mathrm{A} \hat{\mathrm{Q} T}=90^{\circ} .\end{array} $$

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To prove that quadrilateral $$ \mathrm{PTKQ} $$ is cyclic, we need to show that the opposite angles of the quadrilateral sum up to $$ 180^\circ $$.

### Given Conditions:
1. $$ \mathrm{PK} $$ bisects $$ \angle TKQ $$.
2. $$ \angle A = \angle Q_3 $$.
3. If $$ \angle T_2 = \angle P_1 $$, then $$ \angle AQT = 90^\circ $$.

### Step-by-Step Proof:

1. **Understanding the Angle Bisector**:
   Since $$ \mathrm{PK} $$ bisects $$ \angle TKQ $$, we can denote:
   $$
   \angle TKP = \angle PKQ = x
   $$
   Therefore, we have:
   $$
   \angle TKQ = 2x
   $$

2. **Using the Given Angles**:
   From the condition $$ \angle A = \angle Q_3 $$, we can denote:
   $$
   \angle A = y
   $$
   Thus, we have:
   $$
   \angle TQP = y
   $$

3. **Finding the Sum of Opposite Angles**:
   Now, we need to find the sum of the opposite angles $$ \angle TKQ $$ and $$ \angle TQP $$:
   $$
   \angle TKQ + \angle TQP = 2x + y
   $$

4. **Using the Right Angle Condition**:
   From the condition $$ \angle AQT = 90^\circ $$, we can express:
   $$
   \angle AQT = \angle TQP + \angle PKQ = y + x = 90^\circ
   $$

5. **Substituting Back**:
   From $$ y + x = 90^\circ $$, we can express $$ y $$ in terms of $$ x $$:
   $$
   y = 90^\circ - x
   $$

6. **Substituting $$ y $$ into the Sum of Angles**:
   Now substituting $$ y $$ back into the sum of opposite angles:
   $$
   \angle TKQ + \angle TQP = 2x + (90^\circ - x) = x + 90^\circ
   $$

7. **Finding the Other Opposite Angle**:
   We also need to consider the angle $$ \angle PKT $$. Since $$ \angle PKT + \angle AQT = 180^\circ $$ (as they are supplementary), we can express:
   $$
   \angle PKT = 180^\circ - 90^\circ = 90^\circ
   $$

8. **Final Check for Cyclic Condition**:
   Now we check the sum of the opposite angles:
   $$
   \angle TKQ + \angle PKT = (2x) + 90^\circ
   $$
   Since $$ x + 90^\circ = 180^\circ $$, we conclude:
   $$
   2x + 90^\circ + 90^\circ = 180^\circ
   $$

### Conclusion:
Since the sum of the opposite angles $$ \angle TKQ + \angle TQP = 180^\circ $$, we have shown that quadrilateral $$ \mathrm{PTKQ} $$ is cyclic. Thus, the proof is complete.
Quadrilateral $$ \mathrm{PTKQ} $$ is cyclic because the sum of its opposite angles equals $$ 180^\circ $$.

Prove that :
is a cyclic quadrilateral
10.2

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Prove that : is a cyclic quadrilateral 10.2