In triangle $ABC$ with $\overline{AB}=\overline{AC}=3.6$, a point $D$ is taken on $AB$ at a distance $1.2$ from $A$.  Point $D$ is joined to $E$ in the prolongation of $AC$ so that triangle $AED$ is equal in area to $ABC$.  Then $\overline{AE}$ is:
$ \textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6 $
1. Given triangle \(ABC\) with \(\overline{AB} = \overline{AC} = 3.6\), point \(D\) is on \(AB\) such that \(AD = 1.2\). We need to find \(\overline{AE}\) such that the area of triangle \(AED\) is equal to the area of triangle \(ABC\).

2. Draw perpendiculars from \(D\) and \(B\) to \(AC\), meeting \(AC\) at points \(K\) and \(H\) respectively. Since \(D\) is on \(AB\) and \(AD = 1.2\), we have:
   \[
   \frac{AD}{AB} = \frac{1.2}{3.6} = \frac{1}{3}
   \]
   Therefore, the ratio of the lengths of the perpendiculars from \(D\) and \(B\) to \(AC\) is:
   \[
   \frac{DK}{BH} = \frac{1}{3}
   \]

3. The area of triangle \(ABC\) can be expressed as:
   \[
   [ABC] = \frac{1}{2} \cdot AC \cdot BH
   \]
   Similarly, the area of triangle \(ADE\) can be expressed as:
   \[
   [ADE] = \frac{1}{2} \cdot AE \cdot DK
   \]

4. Given that the areas of triangles \(ABC\) and \(ADE\) are equal, we have:
   \[
   [ABC] = [ADE]
   \]
   Substituting the expressions for the areas, we get:
   \[
   \frac{1}{2} \cdot AC \cdot BH = \frac{1}{2} \cdot AE \cdot DK
   \]

5. Simplifying the equation, we get:
   \[
   AC \cdot BH = AE \cdot DK
   \]
   Using the ratio \(\frac{DK}{BH} = \frac{1}{3}\), we substitute \(DK = \frac{1}{3} BH\) into the equation:
   \[
   AC \cdot BH = AE \cdot \left(\frac{1}{3} BH\right)
   \]
   Simplifying further, we get:
   \[
   AC = \frac{1}{3} AE
   \]
   Therefore:
   \[
   AE = 3 \cdot AC
   \]

6. Given \(AC = 3.6\), we find:
   \[
   AE = 3 \cdot 3.6 = 10.8
   \]

The final answer is \(\boxed{10.8}\).