53. Point $K$ lies on the base $AD$ of trapezoid $ABCD$, such that $|AK|=\lambda|AD|$. Find the ratio $|AM|:|AD|$, where $M$ is the point of intersection with $AD$ of the line passing through the points of intersection of the lines $AB$ and $CD$ and the lines $BK$ and $AC$.

Taking $\lambda=1 / n, n=1,2,3, \ldots$, obtain a method for dividing a given segment into $n$ equal parts using a single ruler, if a line parallel to it is given.
53. Let $P$ and $Q$ be the points of intersection of $B K$ and $A C$, $A B$ and $D C$. The line $Q P$ intersects $A D$ at point $M$ and $B C$ at point $N$. Using the similarity of corresponding triangles, we can write the following equalities:

$$
\frac{|A M|}{|M D|}=\frac{|B N|}{|N C|}=\frac{|M K|}{|A M|}=\frac{|A K|-|A M|}{|A M|}
$$

If $|A M|=x|A D|$, then $\frac{|A M|}{|M D|}=\frac{|A M|}{|A D|-|A M|}=\frac{\boldsymbol{x}}{1-x}, \frac{\boldsymbol{x}}{1-x}=$ $=\frac{\lambda-x}{x}$, from which $x=\frac{\lambda}{\lambda+1}$.

Answer: $\frac{|A M|}{|A D|}=\frac{\lambda}{\lambda+1}$.

If $\lambda=\frac{1}{n}$, then $|A M|=\frac{1}{n+1}|A D|$. Thus, by first taking $K$ coinciding with $D (\lambda=1)$, we get $M_{1}$ as the midpoint of $A B$; by taking $K$ coinciding with $M_{1}$, we find that $M_{2}$ separates $1 / 3$ from $A D$, and so on.