In a triangle ABC, the side BC is extended up to D. Such that CD = AC, if `angleBAD = 109^@ and angle ACB = 72^@` then the value of `angleABC` is

Options

A. `35^@`
B. `60^@`
C. `40^@`
D. `45^@`

Answer

To solve the problem, we will use the properties of triangles and the relationships between angles.

### Step-by-Step Solution:

1. **Identify the Angles Given**:
We have:
- \( \angle BAD = 109^\circ \)
- \( \angle ACB = 72^\circ \)

2. **Determine the Angle \( \angle ADB \)**:
Since \( CD = AC \), triangle \( ACD \) is isosceles. Therefore, \( \angle ACD = \angle ACD \).
In triangle \( ABD \), we can find \( \angle ADB \) using the exterior angle theorem:
\[ \angle ADB = \angle BAD - \angle ACB \]
Substituting the values:
\[ \angle ADB = 109^\circ - 72^\circ = 37^\circ \]

3. **Find \( \angle ABC \)**:
In triangle \( ABD \), the sum of the angles is \( 180^\circ \):
\[ \angle ADB + \angle ABD + \angle BAD = 180^\circ \]
Substituting the known values:
\[ 37^\circ + \angle ABC + 109^\circ = 180^\circ \]
Simplifying this gives:
\[ \angle ABC + 146^\circ = 180^\circ \]
Therefore:
\[ \angle ABC = 180^\circ - 146^\circ = 34^\circ \]

4. **Conclusion**:
Thus, the value of \( \angle ABC \) is \( 34^\circ \).

### Final Answer:
\[ \angle ABC = 34^\circ \]