In the following figure .
(i) if ` angle BAD = 96^(@) , " find " angle BCD and angle BFE. `
(ii) Prove that AD is parallel to PE.
` (##SEL_RKB_ICSE_MAT_X_C17_E02_019_Q01.png" width="80%">
(b) ABCD is a parallelogram. A circle through vertices A and B meets side BC at point P and side AD at point Q . Show that quadrilateral PCDQ is cyclic.

Answer

### Step-by-Step Solution

**Given:**
- \( \angle BAD = 96^\circ \)

**(i) Find \( \angle BCD \) and \( \angle BFE \):**

1. **Identify the cyclic quadrilateral:**
Since points A, B, C, and D lie on the circumference of a circle, quadrilateral ABCD is a cyclic quadrilateral.

2. **Use the property of cyclic quadrilaterals:**
In a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \). Therefore:
\[ \angle BAD + \angle BCD = 180^\circ \]
Substituting the known angle:
\[ 96^\circ + \angle BCD = 180^\circ \]

3. **Solve for \( \angle BCD \):**
\[ \angle BCD = 180^\circ - 96^\circ = 84^\circ \]

4. **Find \( \angle BFE \):**
Now consider the angles around point B. Since \( \angle BCD \) and \( \angle BCE \) form a linear pair:
\[ \angle BCD + \angle BCE = 180^\circ \]
Therefore:
\[ 84^\circ + \angle BCE = 180^\circ \]
Solving for \( \angle BCE \):
\[ \angle BCE = 180^\circ - 84^\circ = 96^\circ \]

5. **Use the cyclic property again for quadrilateral BCEF:**
Since points B, C, E, and F also lie on the circumference of a circle, we have:
\[ \angle BCE + \angle BFE = 180^\circ \]
Substituting the known angle:
\[ 96^\circ + \angle BFE = 180^\circ \]

6. **Solve for \( \angle BFE \):**
\[ \angle BFE = 180^\circ - 96^\circ = 84^\circ \]

**Conclusion for part (i):**
- \( \angle BCD = 84^\circ \)
- \( \angle BFE = 84^\circ \)

---

**(ii) Prove that AD is parallel to FE:**

1. **Sum of angles:**
We have already calculated:
\[ \angle BAD = 96^\circ \quad \text{and} \quad \angle BFE = 84^\circ \]
Now, consider the sum of these angles:
\[ \angle BAD + \angle BFE = 96^\circ + 84^\circ = 180^\circ \]

2. **Conclusion about parallel lines:**
Since the sum of the adjacent angles \( \angle BAD \) and \( \angle BFE \) is \( 180^\circ \), it follows that lines AD and FE are parallel.

**Final Conclusion for part (ii):**
- \( AD \parallel FE \)

---

**(b) Show that quadrilateral PCDQ is cyclic:**

1. **Given:**
ABCD is a parallelogram, and a circle passes through points A and B, intersecting BC at P and AD at Q.

2. **Identify angles:**
Since ABCD is a parallelogram, we know:
\[ \angle A + \angle D = 180^\circ \]
Let \( \angle A = \theta \) and \( \angle D = 180^\circ - \theta \).

3. **Using cyclic properties:**
Since points A, B, P, and Q are on the circle, we have:
\[ \angle APB + \angle AQD = 180^\circ \]

4. **Find opposite angles in quadrilateral PCDQ:**
Since \( \angle PCD + \angle QDC \) are opposite angles in quadrilateral PCDQ, we need to show:
\[ \angle PCD + \angle QDC = 180^\circ \]

5. **Use the property of opposite angles:**
From the properties of the cyclic quadrilateral:
\[ \angle PCD + \angle QDC = \angle A + \angle D = 180^\circ \]

**Conclusion for part (b):**
- Quadrilateral PCDQ is cyclic.

---