156. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Points $M$ and $N$ are taken on segments $A A_{1}$ and $B C_{1}$ such that the line $M N$ intersects the line $B_{1} D$. Find

$$
\frac{\left|B C_{1}\right|}{|B N|}-\frac{|A M|}{\left|A A_{1}\right|}
$$
156. Projecting a cube onto a plane perpendicular to $B_{i} D_{\text {, }}$ we obtain a regular hexagon $A B C C_{1} D_{1} A_{1}$ (Fig. 35) with side

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-095.jpg?height=480&width=415&top_left_y=1241&top_left_x=150)

Fig. $35_{s}$ equal to $\sqrt{\frac{2}{3}} a=b$, where $a$ is the edge of the cube (the equilateral triangle $B C_{\overline{\mathbf{1}}} A_{\mathbf{1}}$ projects into an equal triangle, as the plane $B C_{1} A_{\mathbf{i}}$ is perpendicular to $B_{1} D$). Consider $\triangle K A C_{i}$, in it $|K A|=\left|A C_{\mathbf{i}}\right|=2 b$, the line $N M$ passes through the midpoint of $A C_{\mathbf{i}}$. Let

$$
\frac{|A M|}{\left|A A_{1}\right|}=x
$$

Draw $C_{\overline{1}} L$ parallel to $M N_{9}$, we will have:

$$
\frac{|K N|}{\left|K C_{1}\right|}=\frac{|M L|=|A M|}{|K L|}=\frac{\mid K+x}{2+2 x}
$$

then

$$
\begin{aligned}
\frac{|B N|}{\left|B C_{1}\right|}=\frac{2\left(|K N|-\left|B C_{1}\right|\right)}{\left|K C_{1}\right|} & = \\
& =2 \frac{|K N|}{\left|K C_{1}\right|}-1=\frac{2+x}{1+x}-1=\frac{1}{1+x}
\end{aligned}
$$

Thus,

$$
\frac{\left|B C_{1}\right|}{|B N|}-\frac{|A M|}{\left|A A_{1}\right|}=1+x-x=1
$$