Two circumferences of radius $1$ that do not intersect, $c_1$ and $c_2$, are placed inside an angle whose vertex is $O$. $c_1$ is tangent to one of the rays of the angle, while $c_2$ is tangent to the other ray. One of the common internal tangents of $c_1$ and $c_2$ passes through $O$, and the other one intersects the rays of the angle at points $A$ and $B$, with $AO=BO$. Find the distance of point $A$ to the line $OB$.
1. **Understanding the Geometry**:
   - We have two circles \( c_1 \) and \( c_2 \) with radius 1, placed inside an angle with vertex \( O \).
   - \( c_1 \) is tangent to one ray of the angle, and \( c_2 \) is tangent to the other ray.
   - One of the common internal tangents of \( c_1 \) and \( c_2 \) passes through \( O \).
   - The other common internal tangent intersects the rays of the angle at points \( A \) and \( B \), with \( AO = BO \).

2. **Setting Up the Problem**:
   - Let \( C \) be the point on \( AB \) such that \( OC \) is tangent to both \( c_1 \) and \( c_2 \).
   - Since \( AO = BO \), triangle \( \triangle AOB \) is isosceles with \( AO = BO \).

3. **Using Symmetry and Tangents**:
   - Since \( OC \) is tangent to both circles, \( OC \) is perpendicular to \( AB \) at \( C \).
   - Therefore, \( C \) is the midpoint of \( AB \), and \( \angle ACO = \angle BCO = 90^\circ \).

4. **Applying the Pythagorean Theorem**:
   - Let \( OC = x \), \( AC = y \), and \( AO = z \).
   - Since \( \triangle AOC \) is a right triangle, we have:
     \[
     x^2 + y^2 = z^2
     \]

5. **Using the Given Condition**:
   - The distance from \( O \) to the line \( AB \) is \( x \).
   - The distance from \( A \) to the line \( OB \) is the length of the perpendicular from \( A \) to \( OB \), which is \( \frac{xy}{z} \).

6. **Finding the Values of \( x \), \( y \), and \( z \)**:
   - We need to find \( x \), \( y \), and \( z \) such that \( x + y - z = 2 \) and \( x^2 + y^2 = z^2 \).

7. **Solving the System of Equations**:
   - From \( x + y - z = 2 \), we can express \( z \) as:
     \[
     z = x + y - 2
     \]
   - Substituting \( z \) into the Pythagorean equation:
     \[
     x^2 + y^2 = (x + y - 2)^2
     \]
   - Expanding and simplifying:
     \[
     x^2 + y^2 = x^2 + y^2 + 4 + 2xy - 4x - 4y
     \]
     \[
     0 = 4 + 2xy - 4x - 4y
     \]
     \[
     2xy - 4x - 4y + 4 = 0
     \]
     \[
     xy - 2x - 2y + 2 = 0
     \]
     \[
     (x - 2)(y - 2) = 2
     \]

8. **Solving for \( x \) and \( y \)**:
   - Let \( x - 2 = a \) and \( y - 2 = b \), then \( ab = 2 \).
   - We need to find \( a \) and \( b \) such that \( ab = 2 \).

9. **Finding the Distance**:
   - The distance from \( A \) to the line \( OB \) is:
     \[
     \frac{xy}{z} = \frac{(a+2)(b+2)}{a+b+2}
     \]
   - Since \( ab = 2 \), we have:
     \[
     \frac{(a+2)(b+2)}{a+b+2} = \frac{ab + 2a + 2b + 4}{a + b + 2} = \frac{2 + 2a + 2b + 4}{a + b + 2} = \frac{6 + 2(a + b)}{a + b + 2}
     \]
     \[
     = 2
     \]

The final answer is \(\boxed{2}\).