Triangle $ABC$ is right angled such that $\angle ACB=90^{\circ}$ and $\frac {AC}{BC} = 2$. Let the line parallel to side $AC$ intersects line segments $AB$ and $BC$ in $M$ and $N$ such that $\frac {CN}{BN} = 2$. Let $O$ be the intersection point of lines $CM$ and $AN$. On segment $ON$ lies point $K$ such that $OM+OK=KN$. Let $T$ be the intersection point of angle bisector of $\angle ABC$ and line from $K$ perpendicular to $AN$. Determine value of  $\angle MTB$.
1. Given that triangle \( ABC \) is right-angled at \( C \) and \(\frac{AC}{BC} = 2\). This implies that \( \triangle ABC \) is a 1-2-\(\sqrt{5}\) triangle. Let \( AC = 2x \) and \( BC = x \). By the Pythagorean theorem, \( AB = \sqrt{(2x)^2 + x^2} = \sqrt{5}x \).

2. A line parallel to \( AC \) intersects \( AB \) and \( BC \) at \( M \) and \( N \) respectively, such that \( \frac{CN}{BN} = 2 \). This implies that \( BN = y \) and \( CN = 2y \). Since \( BC = x \), we have \( y + 2y = x \), so \( y = \frac{x}{3} \) and \( CN = \frac{2x}{3} \).

3. Since \( MN \parallel AC \), \( \triangle CMN \sim \triangle CAB \). Therefore, \( \frac{CM}{CA} = \frac{CN}{CB} \). Given \( \frac{CN}{CB} = \frac{\frac{2x}{3}}{x} = \frac{2}{3} \), it follows that \( \frac{CM}{2x} = \frac{2}{3} \), so \( CM = \frac{4x}{3} \).

4. Let \( O \) be the intersection of \( CM \) and \( AN \). Since \( \frac{AC}{BC} = 2 \), \( CM \) bisects \( \angle ACB \). Also, since \( \frac{CN}{BN} = 2 \), \( AN \) bisects \( \angle BAC \). Therefore, \( O \) is the incenter of \( \triangle ABC \).

5. On segment \( ON \), point \( K \) is such that \( OM + OK = KN \). Since \( O \) is the incenter, \( OM \) and \( ON \) are equal radii of the incircle. Therefore, \( K \) is the midpoint of \( ON \).

6. Let \( T \) be the intersection of the angle bisector of \( \angle ABC \) and the line from \( K \) perpendicular to \( AN \). Since \( K \) is the midpoint of \( ON \), the perpendicular from \( K \) to \( AN \) will pass through the midpoint of the arc \( MON \) of the circumcircle of \( \triangle OMN \).

7. Since \( B, M, O, T, N \) are concyclic, \( \angle MTB = 90^\circ \).

The final answer is \(\boxed{90^\circ}\).