This construction copies a line segment ${\displaystyle {\overline {AB}}}$ to a target point T. The construction is based on Book I, prop 2.

The construction

1. Let A be one of the end points of ${\displaystyle {\overline {AB}}}$. Note that we are just giving it a name here. (We could replace A with the other end point B).
   
   
   
   
2. Draw a line segment ${\displaystyle {\overline {AT}}}$
   
   
   
   
3. Construct an equilateral triangle ${\displaystyle \triangle ATD}$ (a triangle that has ${\displaystyle {\overline {AT}}}$ as one of its sides).
   
   
   
   
4. Draw the circle ${\displaystyle \circ A,{\overline {AB}}}$, whose center is A and radius is ${\displaystyle {\overline {AB}}}$.
   
   
   
   
5. Draw a line segment starting from D going through A until it intersects ${\displaystyle \circ A,{\overline {AB}}}$ and let the intersection point be E . Get segments ${\displaystyle {\overline {AE}}}$ and ${\displaystyle {\overline {DE}}}$.
   
   
   
   
6. Draw the circle ${\displaystyle \circ D,{\overline {DE}}}$, whose center is D and radius is ${\displaystyle {\overline {DE}}}$.
   
   
   
   
7. Draw a line segment starting from D going through T until it intersects ${\displaystyle \circ D,{\overline {DE}}}$ and let the intersection point be F. Get segments ${\displaystyle {\overline {TF}}}$ and ${\displaystyle {\overline {DF}}}$.
   

Claim

The segment ${\displaystyle {\overline {TF}}}$ is equal to ${\displaystyle {\overline {AB}}}$ and starts at T.

Proof

1. Segments ${\displaystyle {\overline {AB}}}$ and ${\displaystyle {\overline {AE}}}$ are both from the center of ${\displaystyle \circ A,{\overline {AB}}}$ to its circumference. Therefore they equal to the circle radius and to each other.
   
   
   
   
2. Segments ${\displaystyle {\overline {DE}}}$ and ${\displaystyle {\overline {DF}}}$ are both from the center of ${\displaystyle \circ D,{\overline {DE}}}$ to its circumference. Therefore they equal to the circle radius and to each other.
   
   
   
   
3. ${\displaystyle {\overline {DE}}}$ equals to the sum of its parts ${\displaystyle {\overline {DA}}}$ and ${\displaystyle {\overline {AE}}}$.
   
   
   
   
4. ${\displaystyle {\overline {DF}}}$ equals to the sum of its parts ${\displaystyle {\overline {DT}}}$ and ${\displaystyle {\overline {TF}}}$.
   
   
   
   
5. The segment ${\displaystyle {\overline {DA}}}$ is equal to ${\displaystyle {\overline {DT}}}$ since they are the sides of the equilateral triangle ${\displaystyle \triangle ATD}$.
   
   
   
   
6. Since the sum of segments is equal and two of the summands are equal so are the two other summands ${\displaystyle {\overline {AE}}}$ and ${\displaystyle {\overline {TF}}}$.
   
   
   
   
7. Therefore ${\displaystyle {\overline {AB}}}$ equals ${\displaystyle {\overline {TF}}}$.