# $n$-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 $AA_{1}$, $A_{1}A_{2}$, $A_{2}A_{3}$, …, $A_{n-1}A_{n}$. Draw the line $A_{n}B$ and denote by $B_{1}$, $B_{2}$, …, $B_{n-1}$ the points of $AB$ such that

 $A_{1}B_{1}\;\parallel\;A_{2}B_{2}\;\parallel\;\ldots\;\parallel\;A_{n-1}B_{n-1% }\;\parallel\;A_{n}B$

(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 $AA_{1}B_{1}$, $AA_{2}B_{2}$ and $AA_{3}B$ are equal. Because of the equal angles, the triangle $AA_{2}B_{2}$ is similar to the triangle $AA_{3}B$ with the ratio of similarity $2\!:\!3$. Therefore

 $AB_{2}=\frac{2}{3}AB;\quad B_{2}B=\frac{1}{3}AB.$

Also the triangle $AA_{1}B_{1}$ is similar to the triangle $AA_{3}B$ with the line ratio $1\!:\!3$, whence

 $AB_{1}=\frac{1}{3}AB;\quad B_{1}B_{2}=\frac{1}{3}AB.$

The equations show that the points $B_{1}$ and $B_{2}$ divide the line segment $AB$ in 3 equal segments.

Title $n$-section of line segment with compass and straightedge NsectionOfLineSegmentWithCompassAndStraightedge 2013-03-22 17:24:41 2013-03-22 17:24:41 pahio (2872) pahio (2872) 12 pahio (2872) Algorithm msc 51F99 msc 51M05 msc 51-00 CompassAndStraightedgeConstructionOfParallelLine