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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. 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 $0<b\leq 1$.
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 $\triangle CED\sim\triangle CFG$, 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.
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