# 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.

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

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:

• 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 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