Circles $C_1$ and $C_2$ intersect at points $X$ and $Y$ . Point $A$ is a point on $C_1$ such that the tangent line with respect to $C_1$ passing through $A$ intersects $C_2$ at $B$ and $C$, with $A$ closer to $B$ than $C$, such that $2016 \cdot AB = BC$. Line $XY$ intersects line $AC$ at $D$. If circles $C_1$ and $C_2$ have radii of $20$ and $16$, respectively, find $\sqrt{1+BC/BD}$.
1. **Define Variables and Use Power of a Point Theorem:**
   Let \( y = BD \) and \( x = AB \). By the Power of a Point theorem, we know that the power of point \( D \) with respect to circle \( C_1 \) is equal to the power of point \( D \) with respect to circle \( C_2 \). This gives us the equation:
   \[
   DA^2 = DX \cdot DY = DB \cdot DC
   \]
   Given that \( BC = 2016 \cdot AB \), we can write \( BC = 2016x \).

2. **Express \( DC \) in Terms of \( x \) and \( y \):**
   Since \( D \) lies on line \( AC \), we can express \( DC \) as:
   \[
   DC = DB + BC = y + 2016x
   \]

3. **Substitute into the Power of a Point Equation:**
   Substitute \( DA = x - y \) and \( DC = y + 2016x \) into the Power of a Point equation:
   \[
   (x - y)^2 = y(y + 2016x)
   \]

4. **Simplify the Equation:**
   Expand and simplify the equation:
   \[
   x^2 - 2xy + y^2 = y^2 + 2016xy
   \]
   \[
   x^2 - 2xy + y^2 = y^2 + 2016xy
   \]
   Subtract \( y^2 \) from both sides:
   \[
   x^2 - 2xy = 2016xy
   \]
   Combine like terms:
   \[
   x^2 = 2018xy
   \]

5. **Solve for \( x \) in Terms of \( y \):**
   Since \( x \neq 0 \), we can divide both sides by \( x \):
   \[
   x = 2018y
   \]

6. **Calculate \( \frac{BC}{BD} \):**
   Substitute \( x = 2018y \) into the expression for \( BC \):
   \[
   BC = 2016x = 2016 \cdot 2018y
   \]
   Therefore:
   \[
   \frac{BC}{BD} = \frac{2016 \cdot 2018y}{y} = 2016 \cdot 2018
   \]

7. **Calculate \( \sqrt{1 + \frac{BC}{BD}} \):**
   \[
   \sqrt{1 + \frac{BC}{BD}} = \sqrt{1 + 2016 \cdot 2018}
   \]
   Simplify inside the square root:
   \[
   \sqrt{1 + 2016 \cdot 2018} = \sqrt{1 + 2017^2 - 1} = \sqrt{2017^2} = 2017
   \]

The final answer is \(\boxed{2017}\).