Let $ABCD$ be a quadrilateral such that $m(\widehat{A}) = m(\widehat{D}) = 90^\circ$. Let $M$ be the midpoint of $[DC]$. If $AC\perp BM$, $|DC|=12$, and $|AB|=9$, then what is $|AD|$?

$ 
\textbf{(A)}\ 4
\qquad\textbf{(B)}\ 6
\qquad\textbf{(C)}\ 9
\qquad\textbf{(D)}\ 12
\qquad\textbf{(E)}\ \text{None of the above}
$
1. Given that \(ABCD\) is a quadrilateral with \(m(\widehat{A}) = m(\widehat{D}) = 90^\circ\), we can infer that \(AB\) and \(AD\) are perpendicular to each other, and \(AD\) and \(DC\) are perpendicular to each other.
2. Let \(M\) be the midpoint of \(DC\). Since \(M\) is the midpoint, we have \(|DM| = |MC| = \frac{|DC|}{2} = \frac{12}{2} = 6\).
3. Given that \(AC \perp BM\), we can use the Pythagorean theorem in the right triangles formed by these perpendiculars.
4. Let \(AD = x\). Since \(m(\widehat{A}) = 90^\circ\), triangle \(ABD\) is a right triangle with \(AB = 9\) and \(AD = x\). Therefore, \(BD\) can be calculated using the Pythagorean theorem:
   \[
   BD = \sqrt{AB^2 + AD^2} = \sqrt{9^2 + x^2} = \sqrt{81 + x^2}
   \]
5. Similarly, since \(m(\widehat{D}) = 90^\circ\), triangle \(ADC\) is a right triangle with \(AD = x\) and \(DC = 12\). Therefore, \(AC\) can be calculated using the Pythagorean theorem:
   \[
   AC = \sqrt{AD^2 + DC^2} = \sqrt{x^2 + 12^2} = \sqrt{x^2 + 144}
   \]
6. Since \(M\) is the midpoint of \(DC\), \(BM\) is perpendicular to \(AC\). Therefore, triangle \(BMC\) is a right triangle with \(BM\) as the height from \(B\) to \(AC\).
7. Using the fact that \(AC \perp BM\), we can write the equation for the lengths involving \(BM\):
   \[
   BM^2 + AM^2 = AC^2
   \]
8. Since \(M\) is the midpoint of \(DC\), \(AM\) can be calculated as:
   \[
   AM = \sqrt{AD^2 + DM^2} = \sqrt{x^2 + 6^2} = \sqrt{x^2 + 36}
   \]
9. Substituting the values into the equation \(BM^2 + AM^2 = AC^2\):
   \[
   BM^2 + (x^2 + 36) = x^2 + 144
   \]
10. Simplifying the equation:
    \[
    BM^2 + x^2 + 36 = x^2 + 144
    \]
    \[
    BM^2 + 36 = 144
    \]
    \[
    BM^2 = 108
    \]
11. Since \(BM\) is perpendicular to \(AC\), we can use the Pythagorean theorem to find \(AD\):
    \[
    2AD^2 = 72
    \]
    \[
    AD^2 = 36
    \]
    \[
    AD = \sqrt{36} = 6
    \]

The final answer is \(\boxed{6}\).