The bisector of `angleA` of `DeltaABC` cuts BC at D and the circumcircle of the triangle at E. then

Options

A. AB:AC=BD:DC
B. AD:AC=AE:AB
C. AB:AD=AC:AE
D. AB:AB=AE:AC

Answer

To solve the problem, we need to apply the Angle Bisector Theorem in triangle \( ABC \) where the angle bisector of \( \angle A \) intersects \( BC \) at point \( D \) and the circumcircle of triangle \( ABC \) at point \( E \).

### Step-by-Step Solution:

1. **Understanding the Triangle and Points**:
- We have triangle \( ABC \).
- The angle bisector of \( \angle A \) intersects side \( BC \) at point \( D \).
- The angle bisector also intersects the circumcircle of triangle \( ABC \) at point \( E \).

2. **Applying the Angle Bisector Theorem**:
- According to the Angle Bisector Theorem, if a point \( D \) lies on side \( BC \) such that \( AD \) is the angle bisector of \( \angle A \), then:
\[ \frac{AB}{AC} = \frac{BD}{DC} \]
- This means that the ratio of the lengths of the sides \( AB \) and \( AC \) is equal to the ratio of the segments \( BD \) and \( DC \).

3. **Setting Up the Ratios**:
- Let \( AB = c \), \( AC = b \), \( BD = x \), and \( DC = y \).
- From the Angle Bisector Theorem, we can write:
\[ \frac{c}{b} = \frac{x}{y} \]

4. **Conclusion**:
- The relationship established by the Angle Bisector Theorem is the key result we need to answer the question.
- Therefore, the correct relation based on the options provided (assuming they are in the form of ratios) would be:
\[ \frac{AB}{AC} = \frac{BD}{DC} \]