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 the angle $\mathrm{\angle }CFG$ was copied from the angle $\mathrm{\angle }CED$ and the two triangles share the angle $\mathrm{\angle }DCE$, then the two triangles $\mathrm{△}CED$ and $\mathrm{△}CFG$ are similar;

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

• 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.