Right triangle $ABC$ has its right angle at $A$. A semicircle with center $O$ is inscribed inside triangle $ABC$ with the diameter along $AB$. Let $D$ be the point where the semicircle is tangent to $BC$. If $AD=4$ and $CO=5$, find $\cos\angle{ABC}$.

[asy]
import olympiad;

pair A, B, C, D, O;
A = (1,0);
B = origin;
C = (1,1);
O = incenter(C, B, (1,-1));
draw(A--B--C--cycle);
dot(O);
draw(arc(O, 0.41421356237,0,180));
D = O+0.41421356237*dir(135);
label("$A$",A,SE);
label("$B$",B,SW);
label("$C$",C,NE);
label("$D$",D,NW);
label("$O$",O,S);
[/asy]

$\text{(A) }\frac{\sqrt{5}}{4}\qquad\text{(B) }\frac{3}{5}\qquad\text{(C) }\frac{12}{25}\qquad\text{(D) }\frac{4}{5}\qquad\text{(E) }\frac{2\sqrt{5}}{5}$
1. **Identify the given information and set up the problem:**
   - Right triangle \(ABC\) with right angle at \(A\).
   - Semicircle with center \(O\) inscribed inside triangle \(ABC\) with diameter along \(AB\).
   - \(D\) is the point where the semicircle is tangent to \(BC\).
   - \(AD = 4\) and \(CO = 5\).

2. **Determine the radius \(r\) of the semicircle:**
   - Since the semicircle is inscribed with diameter along \(AB\), the radius \(r\) is half the length of \(AB\).
   - Let \(AB = 2r\).

3. **Use the given information \(AD = 4\) and \(CO = 5\):**
   - Since \(AD\) is the distance from \(A\) to the point of tangency \(D\), and \(AD = 4\), we know that \(AD\) is perpendicular to \(BC\).
   - \(CO\) is the distance from \(C\) to the center \(O\) of the semicircle, and \(CO = 5\).

4. **Use the Pythagorean theorem in triangle \(ACO\):**
   - Since \(A\) is the right angle, \(AC\) and \(AO\) are the legs of the right triangle, and \(CO\) is the hypotenuse.
   - Let \(AC = x\) and \(AO = r\).
   - By the Pythagorean theorem: 
     \[
     x^2 + r^2 = 5^2
     \]
     \[
     x^2 + r^2 = 25
     \]

5. **Determine the relationship between \(r\) and \(AD\):**
   - Since \(AD = 4\) and \(AD\) is perpendicular to \(BC\), \(AD\) is the height from \(A\) to \(BC\).
   - The radius \(r\) is also the distance from \(O\) to \(AB\), so \(r = 4\).

6. **Solve for \(x\) using the Pythagorean theorem:**
   - Substitute \(r = 4\) into the equation:
     \[
     x^2 + 4^2 = 25
     \]
     \[
     x^2 + 16 = 25
     \]
     \[
     x^2 = 9
     \]
     \[
     x = 3
     \]

7. **Determine the length of \(AB\):**
   - Since \(AB = 2r\) and \(r = 4\):
     \[
     AB = 2 \times 4 = 8
     \]

8. **Use the cosine definition in triangle \(ABC\):**
   - In right triangle \(ABC\), \(\cos \angle ABC\) is the ratio of the adjacent side \(AB\) to the hypotenuse \(BC\).
   - Use the Pythagorean theorem to find \(BC\):
     \[
     BC^2 = AB^2 + AC^2
     \]
     \[
     BC^2 = 8^2 + 3^2
     \]
     \[
     BC^2 = 64 + 9
     \]
     \[
     BC^2 = 73
     \]
     \[
     BC = \sqrt{73}
     \]

9. **Calculate \(\cos \angle ABC\):**
   - \(\cos \angle ABC = \frac{AB}{BC}\):
     \[
     \cos \angle ABC = \frac{8}{\sqrt{73}}
     \]

The final answer is \(\boxed{\frac{8}{\sqrt{73}}}\)