Geometry for Elementary School/Copying a line segment

This construction copies a line segment ${\overline {AB}}$ 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 ${\overline {AB}}$. 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 ${\overline {AT}}$
   
   
   
   
3. Construct an equilateral triangle $\triangle ATD$ (a triangle that has ${\overline {AT}}$ as one of its sides).
   
   
   
   
4. Draw the circle $\circ A,{\overline {AB}}$, whose center is A and radius is ${\overline {AB}}$.
   
   
   
   
5. Draw a line segment starting from D going through A until it intersects $\circ A,{\overline {AB}}$ and let the intersection point be E . Get segments ${\overline {AE}}$ and ${\overline {DE}}$.
   
   
   
   
6. Draw the circle $\circ D,{\overline {DE}}$, whose center is D and radius is ${\overline {DE}}$.
   
   
   
   
7. Draw a line segment starting from D going through T until it intersects $\circ D,{\overline {DE}}$ and let the intersection point be F. Get segments ${\overline {TF}}$ and ${\overline {DF}}$.
   

Claim

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

Proof

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