Let $E$ be a point outside the square $ABCD$ such that $m(\widehat{BEC})=90^{\circ}$, $F\in [CE]$, $[AF]\perp [CE]$, $|AB|=25$, and $|BE|=7$. What is $|AF|$?

$ 
\textbf{(A)}\ 29
\qquad\textbf{(B)}\ 30
\qquad\textbf{(C)}\ 31
\qquad\textbf{(D)}\ 32
\qquad\textbf{(E)}\ 33
$
1. **Understanding the Problem:**
   - We have a square \(ABCD\) with side length \(AB = 25\).
   - Point \(E\) is outside the square such that \(\angle BEC = 90^\circ\).
   - Point \(F\) lies on segment \([CE]\) such that \([AF] \perp [CE]\).
   - We need to find the length of \([AF]\).

2. **Using Rotation:**
   - Rotate \(\triangle BEC\) \(90^\circ\) about point \(B\). This rotation will map:
     - Point \(B\) to itself.
     - Point \(C\) to point \(A\) (since \(AB = BC\)).
     - Point \(E\) to some point \(E'\) on line \(AF\).

3. **Applying Pythagoras' Theorem:**
   - Since \(\angle BEC = 90^\circ\), \(\triangle BEC\) is a right triangle.
   - Using the Pythagorean theorem in \(\triangle BEC\):
     \[
     EC = \sqrt{BE^2 + BC^2} = \sqrt{7^2 + 25^2} = \sqrt{49 + 625} = \sqrt{674} = 26
     \]
   - After rotation, \(E'\) lies on \(AF\) such that \(AE' = EC = 26\).

4. **Finding \(E'F\):**
   - Since \(BE = BE' = 7\) and \(E'\) lies on \(AF\), the distance \(E'F\) is the same as \(BE\), which is \(7\).

5. **Calculating \(AF\):**
   - The total length of \(AF\) is the sum of \(AE'\) and \(E'F\):
     \[
     AF = AE' + E'F = 26 + 7 = 33
     \]

The final answer is \(\boxed{33}\).