Geometry for elementary school/Copying a line segment

Introduction

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 $\overline{AT}$
   
   
   
   
3. Construct an equilateral triangle $\mathrm{△}ATD$ (a triangle that $\overline{AT}$ is 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 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 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 $\mathrm{△}ATD$.
   
   
   
   
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 $\overline{TF}$.