In a `Delta ABC`, if the bisector of the angle BAC meets BC in D, then which one of the following is correct?

Options

A. `AB le BD`
B. `angle DBA gt (1)/(2) angle BAC`
C. `Delta ABD` can never be isosceles
D. If `BD = AD`, then `Delta ABD` will be equilateral.

Answer

To solve the problem, we need to analyze the triangle \( \Delta ABC \) where the angle bisector of \( \angle BAC \) meets side \( BC \) at point \( D \). We want to determine which statement about the angles in this triangle is correct.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- We have triangle \( ABC \).
- The angle bisector \( AD \) of \( \angle BAC \) meets side \( BC \) at point \( D \).

2. **Understand the Angle Bisector**:
- Since \( AD \) is the angle bisector, it divides \( \angle BAC \) into two equal parts:
\[ \angle DAC = \angle DAB = \frac{1}{2} \angle BAC \]

3. **Consider the Exterior Angle**:
- The exterior angle at vertex \( D \) (formed by extending side \( BC \) beyond point \( D \)) is given by:
\[ \angle BDA = \angle DAB + \angle ABC \]
- This means that \( \angle BDA \) is equal to the sum of \( \angle DAB \) and \( \angle ABC \).

4. **Apply the Exterior Angle Theorem**:
- According to the exterior angle theorem, the exterior angle \( \angle BDA \) is greater than either of the opposite interior angles \( \angle DAC \) or \( \angle ABC \):
\[ \angle BDA > \angle DAC \]

5. **Substituting the Values**:
- Since \( \angle DAC = \frac{1}{2} \angle BAC \), we can express the relationship:
\[ \angle BDA > \frac{1}{2} \angle BAC \]

6. **Conclusion**:
- Therefore, we can conclude that:
\[ \angle BDA > \frac{1}{2} \angle BAC \]
- Among the options provided, the correct statement is that \( \angle BAD \) (which is equal to \( \angle DAB \)) is greater than \( \frac{1}{2} \angle BAC \).

### Final Answer:
The correct statement is that \( \angle BAD > \frac{1}{2} \angle BAC \).