Construction of 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 Definition I-15 that $$BC = BG$$.

As $$D$$ is the center of circle $$GKL$$, it follows from Definition I-15 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 $$BC$$.

Note
This is Proposition 2 of Book I of Euclid's "The Elements".