Let $ABCD$ be a trapezoid and a tangential quadrilateral such that $AD || BC$ and $|AB|=|CD|$. The incircle touches $[CD]$ at $N$. $[AN]$  and $[BN]$ meet the incircle again at $K$ and $L$, respectively. What is $\dfrac {|AN|}{|AK|} + \dfrac {|BN|}{|BL|}$?

$ 
\textbf{(A)}\ 8
\qquad\textbf{(B)}\ 9
\qquad\textbf{(C)}\ 10
\qquad\textbf{(D)}\ 12
\qquad\textbf{(E)}\ 16
$
1. **Understanding the Problem:**
   We are given a trapezoid \(ABCD\) with \(AD \parallel BC\) and \(|AB| = |CD|\). The quadrilateral is tangential, meaning it has an incircle that touches all four sides. The incircle touches \(CD\) at point \(N\). We need to find the value of \(\dfrac{|AN|}{|AK|} + \dfrac{|BN|}{|BL|}\) where \(K\) and \(L\) are the points where \(AN\) and \(BN\) meet the incircle again, respectively.

2. **Symmetry and Properties of Tangential Quadrilateral:**
   Since \(ABCD\) is a tangential quadrilateral, the sum of the lengths of opposite sides are equal:
   \[
   AB + CD = AD + BC
   \]
   Given \(|AB| = |CD|\), it follows that:
   \[
   2|AB| = AD + BC
   \]
   Let \(AD = 2x\) and \(BC = 2y\). Then:
   \[
   2|AB| = 2x + 2y \implies |AB| = x + y
   \]

3. **Using the Incircle:**
   The incircle touches \(CD\) at \(N\). Let \(DN = x\) and \(CN = y\). Since the incircle is tangent to \(CD\) at \(N\), \(N\) divides \(CD\) into segments proportional to \(AD\) and \(BC\).

4. **Calculating \(\dfrac{|AN|}{|AK|}\) and \(\dfrac{|BN|}{|BL|}\):**
   By the properties of tangential quadrilaterals and the symmetry of the trapezoid, we know that:
   \[
   AK \cdot AN = (AD/2)^2 = x^2 \implies AK = \dfrac{x^2}{AN} \implies \dfrac{AN}{AK} = \dfrac{AN^2}{x^2}
   \]
   Similarly:
   \[
   \dfrac{BN}{BL} = \dfrac{BN^2}{y^2}
   \]

5. **Using Similar Triangles:**
   Let \(M\) be a point on the extension of \(DC\) such that \(C\) is between \(M\) and \(D\) and \(CM = BC/2 = y\). Then \(\triangle MCB \sim \triangle NDA\) and:
   \[
   \dfrac{AN}{BM} = \dfrac{ND}{MC} = \dfrac{x}{y} \implies AN = \dfrac{BM \cdot x}{y}
   \]

6. **Median Property in \(\triangle BMN\):**
   In \(\triangle BMN\), \(BC\) is a median. Using the median property:
   \[
   BM^2 + BN^2 = 2(BC^2 + CN^2) = 2(4y^2 + y^2) = 10y^2
   \]
   Therefore:
   \[
   \dfrac{AN}{AK} + \dfrac{BN}{BL} = \dfrac{BM^2}{y^2} + \dfrac{BN^2}{y^2} = \dfrac{10y^2}{y^2} = 10
   \]

Conclusion:
\[
\boxed{10}
\]