The side `B C` of a triangle `A B C` is bisected at `D ;o` is any point in `A D ,B Oa n dC O` produced meet `A Ca n dA B` in `Ea n dF` respectively and `A D` is produced to `X` so that `D` is the mid-point of `O Xdot` Prove that `A O : A X=A F : A B` and show that `F E B Cdot`

Explanation

To solve the problem, we will follow a structured approach to prove that \( \frac{AO}{AX} = \frac{AF}{AB} \) and that \( FE \parallel BC \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Triangle \( ABC \) with \( D \) as the midpoint of side \( BC \).
- Points \( O \) lies on line segments \( AD \), \( BO \), and \( CO \) extended, meeting \( AC \) and \( AB \) at points \( E \) and \( F \) respectively.
- \( AD \) is extended to \( X \) such that \( D \) is the midpoint of \( OX \).
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