# Geometry for Elementary School/Copying a line segment

 Geometry for Elementary School Constructing equilateral triangle Copying a line segment Copying a triangle

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 $\overline{TF}$.