Construction of Equal Straight Line

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Theorem

At a given point, it is possible to construct a straight line segment of length equal to that of any given straight line segment.

The given point will be an endpoint of the constructed straight line segment.


In the words of Euclid:

To place at a given point a straight line equal to a given straight line.

(The Elements: Book $\text{I}$: Proposition $2$)


Construction

Euclid-I-2.png

Let $A$ be a given point.

Let $BC$ be the given straight line segment.


We draw a line segment from $A$ to $B$ to form the straight line segment $AB$.

We construct an equilateral triangle $\triangle ABD$ on $AB$.

We extend the straight lines $DA$ and $DB$ to $E$ and $F$ respectively.

We construct a circle $CGH$ with center $B$ and radius $BC$.

We construct a circle $GKL$ with center $D$ and radius $DG$.


The line $AL$ is the straight line segment required.


Proof

As $B$ is the center of circle $CGH$, it follows from Book $\text{I}$ Definition $15$: Circle that $BC = BG$.

As $D$ is the center of circle $GKL$, it follows from Book $\text{I}$ Definition $15$: Circle that $DL = DG$.

As $\triangle ABD$ is an equilateral triangle, it follows that $DA = DB$.

Therefore, by Common Notion $3$, $AL = BG$.

As $AL = BG$ and $BC = BG$, it follows from Common Notion $1$ that $AL = BC$.


Therefore, at the given point $A$, the required straight line segment $AL$ has been placed equal in length to $BC$.

$\blacksquare$


Historical Note

This proof is Proposition $2$ of Book $\text{I}$ of Euclid's The Elements.


Sources