Construction of an Equal Straight Line

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.

Construction

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 I Definition 15: Circle that $BC = BG$.

As $D$ is the center of circle $GKL$, it follows from Book 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 is Proposition 2 of Book I of Euclid's The Elements.