A line parallel to the side $AC$ of a triangle $ABC$ with $\angle C = 90$ intersects side $AB$ at $M$ and side $BC$ at $N$, so that $CN/BN = AC/BC = 2/1$. The segments $CM$ and $AN$ meet at $O$. Let $K$ be a point on the segment $ON$ such that $MO+OK = KN$. The bisector of $\angle ABC$ meets the line through $K$ perpendicular to $AN$ at point $T$.
Determine $\angle MTB$.
 
1. **Set up the coordinate system:**
   - Place point \( C \) at the origin: \( C = (0,0) \).
   - Place point \( B \) on the x-axis: \( B = (3,0) \).
   - Place point \( A \) such that \( \angle C = 90^\circ \): \( A = (6,0) \).

2. **Determine the coordinates of points \( N \) and \( M \):**
   - Given \( \frac{CN}{BN} = \frac{AC}{BC} = \frac{2}{1} \), we find \( N \) on \( BC \) such that \( CN = 2 \) and \( BN = 1 \). Thus, \( N = (2,0) \).
   - Since the line parallel to \( AC \) intersects \( AB \) at \( M \) and \( AC \) is vertical, \( M \) must have the same x-coordinate as \( N \) and lie on \( AB \). Therefore, \( M = (2,2) \).

3. **Find the equations of lines \( CM \) and \( AN \):**
   - Line \( CM \) passes through \( (0,0) \) and \( (2,2) \), so its equation is \( y = x \).
   - Line \( AN \) passes through \( (6,0) \) and \( (2,0) \), so its equation is \( y = 6 - 3x \).

4. **Find the coordinates of point \( O \):**
   - Solve the system of equations \( y = x \) and \( y = 6 - 3x \):
     \[
     x = 6 - 3x \implies 4x = 6 \implies x = 1.5
     \]
     \[
     y = 1.5
     \]
     Thus, \( O = (1.5, 1.5) \).

5. **Calculate the distances \( OM \) and \( ON \):**
   - \( OM = \sqrt{(2-1.5)^2 + (2-1.5)^2} = \frac{\sqrt{2}}{2} \).
   - \( ON = \sqrt{(2-1.5)^2 + (0-1.5)^2} = \frac{\sqrt{10}}{2} \).

6. **Determine the coordinates of point \( K \):**
   - Given \( MO + OK = KN \), we find \( KN = \frac{OM + ON}{2} = \frac{\sqrt{2} + \sqrt{10}}{4} \).
   - Solve for \( x \)-coordinate of \( K \) using the distance formula:
     \[
     (x-2)^2 + (6-3x)^2 = \left(\frac{\sqrt{2} + \sqrt{10}}{4}\right)^2
     \]
     Solving this, we get \( x = \frac{35 - \sqrt{5}}{20} \) and \( y = \frac{15 + 3\sqrt{5}}{20} \). Thus, \( K = \left(\frac{35 - \sqrt{5}}{20}, \frac{15 + 3\sqrt{5}}{20}\right) \).

7. **Find the equation of line \( TK \):**
   - The slope of \( TK \) is given by the coordinates of \( K \):
     \[
     y = \frac{2x + 1 + \sqrt{5}}{6}
     \]

8. **Find the coordinates of point \( G \):**
   - \( G \) is the intersection of the angle bisector of \( \angle ABC \) and \( AC \):
     \[
     G = \left(0, \frac{3(\sqrt{5} - 1)}{2}\right)
     \]

9. **Find the equation of the angle bisector \( BG \):**
   - The slope of \( BG \) is:
     \[
     y = \frac{(1 - \sqrt{5})x + 3\sqrt{5} - 3}{2}
     \]

10. **Find the coordinates of point \( T \):**
    - Solve the system of equations for \( BG \) and \( TK \):
      \[
      \frac{(1 - \sqrt{5})x + 3\sqrt{5} - 3}{2} = \frac{2x + 1 + \sqrt{5}}{6}
      \]
      Solving this, we get \( x = \frac{5 - \sqrt{5}}{2} \) and \( y = 1 \). Thus, \( T = \left(\frac{5 - \sqrt{5}}{2}, 1\right) \).

11. **Verify \( \angle MTB = 90^\circ \):**
    - Calculate \( MT^2 + TB^2 - MB^2 \):
      \[
      MT^2 + TB^2 - MB^2 = \frac{(\sqrt{5} - 1)^2}{4} + 1 + \frac{(\sqrt{5} + 1)^2}{4} + 1 - 5 = 0
      \]
    - Since the sum of the squares of the distances equals zero, \( \angle MTB = 90^\circ \).

\(\blacksquare\)

The final answer is \( \boxed{ \angle MTB = 90^\circ } \)