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Let's break down the problem step by step.

We are given:
- \( ABCD \) is a square.
- \( F \) is a point on side \( AB \) such that \( DF \parallel EC \).
- \( EC = ED \).
- \( \angle CED = 48^\circ \).

We need to find:
1. \( \angle ADF \) (part (a)).
2. \( \angle BCF \), correct to the nearest degree (part (b)).

### Key Observations:
- Since \( ABCD \) is a square, all angles inside the square are \( 90^\circ \), and all sides are equal in length.
- \( DF \parallel EC \) means that triangles \( ADF \) and \( CED \) are similar.
- \( EC = ED \) implies triangle \( CED \) is isosceles.

#### Part (a) – Finding \( \angle ADF \):
- Since \( DF \parallel EC \) and \( ABCD \) is a square, \( \angle ADF \) is an alternate interior angle to \( \angle CED \).
- \( \angle CED = 48^\circ \), so \( \angle ADF = 48^\circ \) as well.

#### Part (b) – Finding \( \angle BCF \):
- \( \triangle CED \) is isosceles, with \( EC = ED \) and \( \angle CED = 48^\circ \).
- The sum of the angles in triangle \( CED \) is \( 180^\circ \), so the other two angles \( \angle EDC \) and \( \angle ECD \) are equal.
  
Let each of these angles be \( \theta \). Thus:
\[
\theta + \theta + 48^\circ = 180^\circ
\]
\[
2\theta = 132^\circ
\]
\[
\theta = 66^\circ
\]

So, \( \angle EDC = \angle ECD = 66^\circ \).

Since \( DF \parallel EC \), we can use the property of corresponding angles to conclude:
\[
\angle BCF = \angle EDC = 66^\circ
\]

Thus, \( \angle BCF = 66^\circ \).

### Final Answers:
- (a) \( \angle ADF = 48^\circ \)
- (b) \( \angle BCF = 66^\circ \)

Would you like further clarification on any part? Here are 5 related questions to expand your understanding:

1. Why are triangles \( ADF \) and \( CED \) considered similar?
2. What other properties of isosceles triangles can be used to solve this problem?
3. How do parallel lines affect the angle relationships in this diagram?
4. Can you explain why \( \angle ADF \) is equal to \( \angle CED \)?
5. How does the square's symmetry simplify finding unknown angles?

**Tip**: Always look for geometric properties like parallel lines and symmetry, as they often simplify complex diagrams.

In the figure, ABCD is a square. F is a point lying on AB such that DF // EC. It is given that EC = ED and ∠CED = 48°. Find ∠ADF and ∠BCF, correct to the nearest degree.

This geometry problem involves a square, parallel lines, and an isosceles triangle. The problem asks to find ∠ADF and ∠BCF, applying the properties of alternate interior angles, corresponding angles, and the isosceles triangle theorem. Suitable for students in Grades 9-10.

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