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 $0<b\le 1$ .

   
   
   
6. Connect the points $D$ and $E$ .
   
   
   
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 $0<b<1$ , 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 the angle $\angle CFG$ was copied from the angle $\angle CED$ and the two triangles share the angle $\angle DCE$ , then the two triangles $\triangle CED$ and $\triangle CFG$ are similar;
• Since $\triangle CED \sim \triangle CFG$ , we have that $\displaystyle \frac{CE}{CF}=\frac{CD}{CG}$ ;
• 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 link provided.