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# $n$-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.

Solution. 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 $A_{1}B_{1}A$, $A_{2}B_{2}A$ 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.

## Mathematics Subject Classification

51F99*no label found*51M05

*no label found*51-00

*no label found*

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