In the given figure The bisector of interior `/_A` of ` A B C` meets `B C` in `D ,` and the bisector of exterior `/_A` meets `B C` produced in `Edot` Prove that `(B D)/(B E)=(C D)/(C E)dot`

Explanation

To prove that \(\frac{BD}{BE} = \frac{CD}{CE}\), we will use the properties of angle bisectors in triangles.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Let \(AD\) be the internal angle bisector of \(\angle A\) in triangle \(ABC\), meeting \(BC\) at point \(D\).
- Let \(AE\) be the external angle bisector of \(\angle A\), meeting the extended line \(BC\) at point \(E\).

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