In triangle $ABC,$ $\sin \angle A=\frac{4}{5}$ and $\angle A<90^\circ$  Let $D$ be a point outside triangle $ABC$ such that $\angle BAD=\angle DAC$ and $\angle BDC = 90^{\circ}.$ Suppose that $AD=1$ and that $\frac{BD} {CD} = \frac{3}{2}.$ If $AB+AC$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ where $a,b,c$ are pairwise relatively prime integers, find $a+b+c$.

[i]Author: Ray Li[/i]
1. First, we need to compute \(\sin \angle BAD = \sin \angle DAC = \sin \frac{A}{2}\). Using the half-angle formula for sine, we have:
   \[
   \sin \frac{A}{2} = \sqrt{\frac{1 - \cos A}{2}}
   \]
   Given \(\sin A = \frac{4}{5}\), we can find \(\cos A\) using the Pythagorean identity:
   \[
   \cos^2 A + \sin^2 A = 1 \implies \cos^2 A + \left(\frac{4}{5}\right)^2 = 1 \implies \cos^2 A = 1 - \frac{16}{25} = \frac{9}{25} \implies \cos A = \frac{3}{5}
   \]
   Therefore,
   \[
   \sin \frac{A}{2} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{\frac{2}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}
   \]

2. We use Cartesian coordinates and scale the diagram suitably. Set \(D = (0,0)\), \(B = (3,0)\), and \(C = (0,2)\). We will scale back at the end so that \(AD = 1\).

3. Define \(P = (3,6)\) and \(Q = (4,2)\). Notice that \(\sin \angle DPB = \sin \angle CQD = \frac{1}{\sqrt{5}}\), so \(CDQA\) and \(BDPA\) are cyclic. Thus, \(A\) is the intersection of the circumcircles of \(\triangle CDQ\) and \(\triangle BDP\) (other than \(D\)).

4. Since \(\triangle CDQ\) and \(\triangle BDP\) are right triangles, finding the equations of their circumcircles is straightforward. The circumcircle of a right triangle with the right angle at the origin has its center at the midpoint of the hypotenuse. Therefore, we get:
   \[
   \left\{
   \begin{aligned}
   (CDQ): & \; (x-2)^2 + (y-1)^2 = 5 \\
   (BDP): & \; \left(x - \frac{3}{2}\right)^2 + (y - 3)^2 = \frac{45}{4}
   \end{aligned}
   \right.
   \]

5. Solving this system of equations, we find:
   \[
   A = (x, y) = \left(\frac{72}{17}, \frac{18}{17}\right)
   \]

6. Next, we compute the distances \(AB\), \(AC\), and \(AD\):
   \[
   AB = \sqrt{\left(3 - \frac{72}{17}\right)^2 + \left(0 - \frac{18}{17}\right)^2} = \frac{3\sqrt{85}}{17}
   \]
   \[
   AC = \sqrt{\left(0 - \frac{72}{17}\right)^2 + \left(2 - \frac{18}{17}\right)^2} = \frac{8\sqrt{85}}{17}
   \]
   \[
   AD = \sqrt{\left(0 - \frac{72}{17}\right)^2 + \left(0 - \frac{18}{17}\right)^2} = \frac{18\sqrt{17}}{17}
   \]

7. Scaling back so that \(AD = 1\), we get the true lengths:
   \[
   AB = \frac{\frac{3\sqrt{85}}{17}}{\frac{18\sqrt{17}}{17}} = \frac{3\sqrt{5}}{18}
   \]
   \[
   AC = \frac{\frac{8\sqrt{85}}{17}}{\frac{18\sqrt{17}}{17}} = \frac{4\sqrt{5}}{9}
   \]

8. Therefore, the sum \(AB + AC\) is:
   \[
   AB + AC = \frac{3\sqrt{5}}{18} + \frac{4\sqrt{5}}{9} = \frac{3\sqrt{5}}{18} + \frac{8\sqrt{5}}{18} = \frac{11\sqrt{5}}{18}
   \]

The final answer is \( \boxed{ a + b + c = 11 + 5 + 18 = 34 } \).