# 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 ${\overline {TF}}$ . 