144. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}, \quad M$ is the center of the face $A B B_{1} A_{1}, N$ is a point on the edge $B_{1} C_{1}, L$ is the midpoint of $A_{1} B_{1}$; $K$ is the foot of the perpendicular dropped from $N$ to $B C_{1}$. In what ratio does the point $N$ divide the edge $B_{1} C_{1}$, if $\widehat{L M K}=\widehat{M K N}$?
144. Let the edge of the cube be $a, |N C_{1}|=x$. Find:

$$
\begin{array}{r}
|L M|=\frac{a}{2}, \quad|N K|=\frac{x}{\sqrt{2}} \\
|L N|^{2}=\left|L B_{1}\right|^{2}+\left|B_{1} N\right|^{2}=\frac{a^{2}}{4}+(a-x)^{2}=\frac{5}{4} a^{2}-2 a x+x^{2} \\
|L K|^{2}=\left|L B_{1}\right|^{2}+\left|B_{1} K\right|^{2}=\left|L B_{1}\right|^{2}+\left|B_{1} N\right|^{2}+|N K|^{2}+ \\
+2\left|B_{1} N\right| \cdot|N K| \frac{\sqrt{2}}{2}=\frac{a^{2}}{4}+(a-x)^{2}+\frac{x^{2}}{2}+(a-x) x= \\
=\frac{5}{4} a^{2}-a x+\frac{x^{2}}{2} \\
|M K|^{2}=|M B|^{2}+|B K|^{2}-|M B| \cdot|B K|^{2}=\frac{3 a^{2}}{2}-\frac{3}{2} a x+\frac{x^{2}}{2}
\end{array}
$$

If $\widehat{L M K}=\widehat{M K N}=\varphi$, then by the cosine theorem for triangles $L M K$ and $M K N$ we get:

$$
\begin{aligned}
|L K|^{2} & =|L M|^{2}+|M K|^{2}-2|L M| \cdot|M K| \cos \varphi \\
|M N|^{2} & =|M K|^{2}+|K N|^{2}-2|M K| \cdot|K N| \cos \varphi
\end{aligned}
$$

Eliminating $\cos \varphi$ from these equations, we get

$$
\begin{aligned}
& |L K|^{2} \cdot|K N|-|M N|^{2} \cdot|L M|= \\
& \quad=(|L M|-|K N|)\left(|L M| \cdot|K N|-|M K|^{2}\right)
\end{aligned}
$$

Substituting the segments involved in this equation using the formulas found above, we get

$$
\begin{aligned}
\left(\frac{5 a^{2}}{4}-a x+\frac{x^{2}}{2}\right) & \frac{x}{\sqrt{2}}-\left(\frac{3 a^{2}}{2}-2 a x+x^{2}\right) \frac{a}{2}= \\
& =\left(\frac{a}{2}-\frac{x}{\sqrt{2}}\right)\left(\frac{a x}{2 \sqrt{2}}-\frac{3 a^{2}}{2}+\frac{3 a x}{2}-\frac{x^{2}}{2}\right)
\end{aligned}
$$

From this equation, we find $x=a\left(1-\frac{\sqrt{2}}{2}\right)$.

Answer: $\frac{\left|B_{1} N\right|}{\left|N C_{1}\right|}=\sqrt{2}+1$.