$n$-section of line segment with compass and straightedge

Task. Let $AB$ be a given line segment and $n$ a positive integer $>1$. Divide $AB$ to $n$ equal parts.

. Draw a half-line $p$ beginning from $A$ but not parallel to $AB$. From $p$ separate $n$ consecutive equally long segments $A{A}_1$, $A_1{A}_2$, $A_2{A}_3$, …, $A_{n-1}{A}_n$. Draw the line $A_{n}B$ and denote by $B_1$, $B_2$, …, $B_{n-1}$ the points of $AB$ such that

$$A_1{B}_1\parallel A_2{B}_2\parallel \mathrm{\dots }\parallel A_{n-1}{B}_{n-1}\parallel A_{n}B$$

(see compass and straightedge construction of parallel line).These points divide the line segment $AB$ in $n$ equal segments.

Proof. For clarity, we prove the theorem only in the case  $n=3$.

The line $AB$ intersects the parallel lines $A_1{B}_1$, $A_2{B}_2$ and $A_{3}B$, and thus the corresponding angles (http://planetmath.org/CorrespondingAnglesInTransversalCutting) $A_1{B}_{1}A$, $A_2{B}_{2}A$ and $A_{3}BA$ are equal. Similarly the angles $A{A}_1{B}_1$, $A{A}_2{B}_2$ and $A{A}_{3}B$ are equal. Because of the equal angles, the triangle $A{A}_2{B}_2$ is similar to the triangle $A{A}_{3}B$ with the ratio of similarity $2:3$. Therefore

$$A{B}_2=\frac{2}{3}AB;B_{2}B=\frac{1}{3}AB.$$

Also the triangle $A{A}_1{B}_1$ is similar to the triangle $A{A}_{3}B$ with the line ratio $1:3$, whence

$$A{B}_1=\frac{1}{3}AB;B_1{B}_2=\frac{1}{3}AB.$$

The equations show that the points $B_1$ and $B_2$ divide the line segment $AB$ in 3 equal segments.