In a triangle ABC, angle bisector of `angle BAC ` cut the side BC at D and meet the circumcircle of `Delta ABC` at E , then find AB .AC+DE. AE.

Options

A. 2AE
B. `AE^(2)`
C. `2AE^(2)`
D. `3AE^(2)`

Answer

To solve the problem, we need to find the expression \( AB \cdot AC + DE \cdot AE \) in triangle \( ABC \) where \( D \) is the point where the angle bisector of \( \angle BAC \) intersects side \( BC \), and \( E \) is the point where the angle bisector meets the circumcircle of triangle \( ABC \).

### Step-by-Step Solution:

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

2. **Using the Angle Bisector Theorem**:
- According to the angle bisector theorem, we have:
\[ \frac{AB}{AC} = \frac{BD}{DC} \]
- This theorem will help us relate the sides of the triangle to the segments created by the angle bisector.

3. **Establishing Similar Triangles**:
- Consider triangles \( ABE \) and \( ACD \). Since \( E \) lies on the circumcircle and \( D \) is on \( BC \), we can establish that:
- \( \angle ABE = \angle ACD \) (angles subtended by the same arc).
- \( \angle AEB = \angle ADC \) (angles subtended by the same arc).
- Therefore, triangles \( ABE \) and \( ACD \) are similar.

4. **Setting Up Proportions**:
- From the similarity of triangles \( ABE \) and \( ACD \), we can write:
\[ \frac{AB}{AC} = \frac{AE}{AD} \]
- Rearranging gives us:
\[ AB \cdot AD = AC \cdot AE \]

5. **Finding \( AB \cdot AC + DE \cdot AE \)**:
- We know from the properties of the angle bisector that:
\[ AD + DE = AE \]
- Thus, we can express \( DE \) as:
\[ DE = AE - AD \]
- Now substituting \( DE \) in our expression:
\[ DE \cdot AE = (AE - AD) \cdot AE = AE^2 - AD \cdot AE \]

6. **Combining the Expressions**:
- Now, substituting back into our original expression:
\[ AB \cdot AC + DE \cdot AE = AB \cdot AC + (AE^2 - AD \cdot AE) \]
- From the earlier similarity, we know \( AB \cdot AD = AC \cdot AE \), so we can replace \( AD \cdot AE \):
\[ AB \cdot AC + DE \cdot AE = AB \cdot AC + AE^2 - AC \cdot AE \]
- Rearranging gives:
\[ AB \cdot AC + DE \cdot AE = AE^2 \]

### Final Result:
Thus, we conclude that:
\[ AB \cdot AC + DE \cdot AE = AE^2 \]