42. $A B C A_{1} B_{1} C_{1}$ - a regular triangular prism, all edges of which are equal to each other. $K$ is a point on edge $A B$, different from $A$ and $B, M$ is on line $B_{1} C_{1}, L$ is in the plane of face $A C C_{1} A_{1}$. Line $K L$ forms equal angles with planes $A B C$ and $A B B_{1} A_{1}, L M$ forms equal angles with planes $B C C_{1} B_{1}$ and $A C C_{1} A_{1}, K M$ also forms equal angles with planes $B C C_{1} B_{1}$ and $A C C_{1} A_{1}$. It is known that $|K L|=|K M|=1$. Find the edge of the prism.
42. Let the side of the base and the height of the prism be denoted by $a$, $|K B|=x$. From the problem statement, it follows that the projection of $K M$ onto the base plane is parallel to the bisector of angle $C$ of triangle $A B C$, i.e., $\left|B_{1} M\right|=2 x,\left|M C_{1}\right|=a-2 x$. Let $L_{1}$ be the projection of $L$ onto $A C$. From the problem statement, we can also obtain that $\left|L L_{1}\right|=$ $=\left|A L_{1}\right| \frac{\sqrt{3}}{2},\left|L_{1} C\right|=a-2 x$. Therefore, the value of $\left|A L_{1}\right|$ can be: 1) $\left|A L_{1}\right|=a-\left|M C_{1}\right|=a-$ $-(a-2 x)=2 x ; 2)\left|A L_{1}\right|=a+(a-2 x)=2(a-x)$. In the first case, $|K L|^{2}=\left|K L_{1}\right|^{2}+\left|L L_{1}\right|^{2}=a^{2}+10 x^{2}-4 a x$; in the second case, $|\bar{K} L|^{2}=6(a-x)^{2}$.

In both cases, $|K M|^{2}=3 x^{2}+a^{2}$.

Solving the respective two systems of equations, we obtain two values for $a$

$$
a_{1}=\frac{7}{\sqrt{97}}, \quad a_{2}=\frac{\sqrt{6}+\sqrt{14}}{8}
$$

Answer: $\frac{7}{\sqrt{97}}, \frac{\sqrt{6}+\sqrt{14}}{8}$.