# 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 ${\overline {TF}}$ .