Let $K$ be a point on the side $[AB]$, and $L$ be a point on the side $[BC]$ of the square $ABCD$. If $|AK|=3$, $|KB|=2$, and the distance of $K$ to the line $DL$ is $3$, what is $|BL|:|LC|$?

$ \textbf{(A)}\ \frac78 
\qquad\textbf{(B)}\ \frac{\sqrt 3}2
\qquad\textbf{(C)}\ \frac 87
\qquad\textbf{(D)}\ \frac 38
\qquad\textbf{(E)}\ \frac{\sqrt 2}2
$
1. **Identify the given information and set up the problem:**
   - \(ABCD\) is a square with side length \(5\).
   - \(K\) is a point on \(AB\) such that \(|AK| = 3\) and \(|KB| = 2\).
   - \(L\) is a point on \(BC\) such that the distance from \(K\) to the line \(DL\) is \(3\).
   - We need to find the ratio \(|BL| : |LC|\).

2. **Place the square in a coordinate system:**
   - Let \(A = (0, 0)\), \(B = (5, 0)\), \(C = (5, 5)\), and \(D = (0, 5)\).
   - Then, \(K = (3, 0)\) since \(|AK| = 3\).

3. **Determine the coordinates of \(L\):**
   - Let \(L = (5, y)\) where \(0 \leq y \leq 5\).
   - We need to find \(y\) such that the distance from \(K\) to the line \(DL\) is \(3\).

4. **Find the equation of the line \(DL\):**
   - The line \(DL\) passes through \(D = (0, 5)\) and \(L = (5, y)\).
   - The slope of \(DL\) is \(\frac{y - 5}{5 - 0} = \frac{y - 5}{5}\).
   - The equation of the line \(DL\) in point-slope form is:
     \[
     y - 5 = \frac{y - 5}{5}x \implies y = \frac{y - 5}{5}x + 5
     \]

5. **Calculate the distance from \(K\) to the line \(DL\):**
   - The distance from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:
     \[
     \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
     \]
   - Rewrite the line equation \(y = \frac{y - 5}{5}x + 5\) in standard form:
     \[
     \frac{y - 5}{5}x - y + 5 = 0 \implies (y - 5)x - 5y + 25 = 0
     \]
   - For \(K = (3, 0)\), the distance is:
     \[
     \frac{|(y - 5) \cdot 3 - 5 \cdot 0 + 25|}{\sqrt{(y - 5)^2 + (-5)^2}} = 3
     \]
     \[
     \frac{|3y - 15 + 25|}{\sqrt{(y - 5)^2 + 25}} = 3
     \]
     \[
     \frac{|3y + 10|}{\sqrt{(y - 5)^2 + 25}} = 3
     \]

6. **Solve for \(y\):**
   - Square both sides to eliminate the square root:
     \[
     \left(\frac{|3y + 10|}{\sqrt{(y - 5)^2 + 25}}\right)^2 = 3^2
     \]
     \[
     \frac{(3y + 10)^2}{(y - 5)^2 + 25} = 9
     \]
     \[
     (3y + 10)^2 = 9((y - 5)^2 + 25)
     \]
     \[
     9y^2 + 60y + 100 = 9(y^2 - 10y + 25 + 25)
     \]
     \[
     9y^2 + 60y + 100 = 9y^2 - 90y + 450
     \]
     \[
     60y + 100 = -90y + 450
     \]
     \[
     150y = 350
     \]
     \[
     y = \frac{350}{150} = \frac{7}{3}
     \]

7. **Determine the lengths \(|BL|\) and \(|LC|\):**
   - \(|BL| = 5 - y = 5 - \frac{7}{3} = \frac{15}{3} - \frac{7}{3} = \frac{8}{3}\)
   - \(|LC| = y = \frac{7}{3}\)

8. **Find the ratio \(|BL| : |LC|\):**
   - \[
     \frac{|BL|}{|LC|} = \frac{\frac{8}{3}}{\frac{7}{3}} = \frac{8}{7}
     \]

The final answer is \(\boxed{\frac{8}{7}}\)