ABC is a triangle in which AB=AC. Let BC be produced to D . From a point E on the line AC. Let EF be a straight line such that EF is parallel to AB. Consider the quadrilateral ECDF thus fromed . If `angle ABC =65^(@)` and `angleEFD=50^(@)`, then find `angleFDC` .

Answer

To solve the problem step by step, we will analyze the given information and use the properties of triangles and parallel lines.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Triangle ABC is isosceles with \( AB = AC \).
- \( \angle ABC = 65^\circ \)
- \( \angle EFD = 50^\circ \)
- Line EF is parallel to line AB.

2. **Determine the Angles in Triangle ABC:**
- Since \( AB = AC \), the angles opposite to these sides are equal. Therefore, \( \angle ACB = \angle ABC = 65^\circ \).
- The sum of angles in triangle ABC is \( 180^\circ \):
\[ \angle A + \angle ABC + \angle ACB = 180^\circ \]
\[ \angle A + 65^\circ + 65^\circ = 180^\circ \]
\[ \angle A + 130^\circ = 180^\circ \]
\[ \angle A = 50^\circ \]

3. **Analyze the Quadrilateral ECDF:**
- Since EF is parallel to AB, by the Alternate Interior Angles Theorem, \( \angle EFD = \angle ABC = 65^\circ \).
- We already know \( \angle EFD = 50^\circ \).

4. **Find Angle DCE:**
- Since line BC is extended to D, \( \angle DCE \) can be found as follows:
\[ \angle DCE = 180^\circ - \angle ABC = 180^\circ - 65^\circ = 115^\circ \]

5. **Use the Angles in Quadrilateral ECDF:**
- The sum of the angles in quadrilateral ECDF is \( 360^\circ \):
\[ \angle EFD + \angle DCE + \angle FDC + \angle CEF = 360^\circ \]
- We know:
- \( \angle EFD = 50^\circ \)
- \( \angle DCE = 115^\circ \)
- \( \angle CEF = 130^\circ \) (as calculated from the parallel lines)
- Plugging in the known values:
\[ 50^\circ + 115^\circ + \angle FDC + 130^\circ = 360^\circ \]
\[ 295^\circ + \angle FDC = 360^\circ \]
\[ \angle FDC = 360^\circ - 295^\circ = 65^\circ \]

### Final Answer:
\[ \angle FDC = 65^\circ \]