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

2.
Extend the line segment past both $C$ and $D$

3.
Erect the perpendicular^{} to $\overleftrightarrow{CD}$ at $C$.

4.
Use the compass to determine a point $E$ on the erected perpendicular such that $CE=1$.

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 $$.

6.
Connect the points $D$ and $E$.

7.
Copy the angle $\mathrm{\angle}CDE$ at $F$ to form similar triangles^{}. Label the intersection^{} of the constructed ray and $\overleftrightarrow{CD}$ as $G$.
Note that, if $$, 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:
 •

•
Since $\mathrm{\u25b3}CED\sim \mathrm{\u25b3}CFG$, we have that $\frac{CE}{CF}}={\displaystyle \frac{CD}{CG}$;

•
Plugging in $CD=a$, $CE=1$, and $CF=b$ yields that $CG=ab$.
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 

Canonical name  CompassAndStraightedgeConstructionOfSimilarTriangles 
Date of creation  20130322 17:16:02 
Last modified on  20130322 17:16:02 
Owner  Wkbj79 (1863) 
Last modified by  Wkbj79 (1863) 
Numerical id  7 
Author  Wkbj79 (1863) 
Entry type  Algorithm^{} 
Classification  msc 51M15 
Classification  msc 51F99 