# compass and straightedge construction of similar triangles

Let $a>0$ and $b>0$. If line segments  of lengths $a$ and $b$ are constructible  , one can construct a line segment of length $ab$ using compass and straightedge as follows:

1. 1.

Draw a line segment of length $a$. Label its endpoints  as $C$ and $D$.

2. 2.

Extend the line segment past both $C$ and $D$

3. 3.
4. 4.

Use the compass to determine a point $E$ on the erected perpendicular such that $CE=1$.

5. 5.

Use the compass to determine a point $F$ on $\overrightarrow{CE}$ such that $CF=b$.

Note that the pictures indicate that $b>1$, but the exact same procedure works if $0.

6. 6.

Connect the points $D$ and $E$.

7. 7.

Copy the angle $\angle CDE$ at $F$ to form similar triangles  . Label the intersection  of the constructed ray and $\overleftrightarrow{CD}$ as $G$.

Note that, if $0, then $F$ will be between $C$ and $E$, and $G$ will be between $C$ and $D$. Also, if $b=1$, then $F=E$ and $G=D$.

This construction is justified by the following:

If you are interested in seeing the rules for compass and straightedge constructions, click on the provided.

Title compass and straightedge construction of similar triangles CompassAndStraightedgeConstructionOfSimilarTriangles 2013-03-22 17:16:02 2013-03-22 17:16:02 Wkbj79 (1863) Wkbj79 (1863) 7 Wkbj79 (1863) Algorithm  msc 51M15 msc 51F99