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[parent] compass and straightedge construction of similar triangles (Algorithm)

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 $D$ .

    \begin{pspicture}(-1,-1)(6,1) \rput[a](0,0.08){.} \psline[linecolor=blue](0,0)(4... ...\psdots(0,0)(4,0) \rput[a](-0.4,-0.3){$C$} \rput[a](4,-0.3){$D$} \end{pspicture}
  2. Extend the line segment past both $C$ and $D$

    \begin{pspicture}(-1,-1)(6,1) \rput[a](0,0.08){.} \rput[l](-1,0){.} \rput[r](6,0... ...\psdots(0,0)(4,0) \rput[a](-0.4,-0.3){$C$} \rput[a](4,-0.3){$D$} \end{pspicture}
  3. Erect the perpendicular to $\overleftrightarrow{CD}$ at $C$ .

    \begin{pspicture}(-1,-1)(6,3) \rput[l](-1,0){.} \rput[r](6,0){.} \rput[a](0,3){.... ...\psdots(0,0)(4,0) \rput[a](-0.4,-0.3){$C$} \rput[a](4,-0.3){$D$} \end{pspicture}
  4. Use the compass to determine a point $E$ on the erected perpendicular such that $CE=1$ .

    \begin{pspicture}(-1,-1)(6,3) \rput[l](-1,0){.} \rput[r](6,0){.} \rput[a](0,3){.... ...a](-0.4,-0.3){$C$} \rput[a](4,-0.3){$D$} \rput[r](-0.4,1.8){$E$} \end{pspicture}
  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 $0<b\le 1$ .


    \begin{pspicture}(-1,-1)(6,3) \rput[l](-1,0){.} \rput[r](6,0){.} \rput[a](0,3){.... ...[a](4,-0.3){$D$} \rput[r](-0.4,1.8){$E$} \rput[r](-0.4,2.3){$F$} \end{pspicture}
  6. Connect the points $D$ and $E$ .

    \begin{pspicture}(-1,-1)(6,3) \rput[l](-1,0){.} \rput[r](6,0){.} \rput[a](0,3){.... ...[a](4,-0.3){$D$} \rput[r](-0.4,1.8){$E$} \rput[r](-0.4,2.3){$F$} \end{pspicture}
  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 $G=D$ .


    \begin{pspicture}(-1,-1)(6,3) \rput[l](-1,0){.} \rput[r](6,0){.} \rput[a](0,3){.... ...[r](-0.4,1.8){$E$} \rput[r](-0.4,2.3){$F$} \rput[a](5,-0.3){$G$} \end{pspicture}

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.




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Cross-references: compass and straightedge constructions, similar, triangles, angle, ray, intersection, similar triangles, copy the angle, point, erect the perpendicular, endpoints, straightedge, compass, constructible, lengths, line segments
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This is version 4 of compass and straightedge construction of similar triangles, born on 2007-06-16, modified 2007-06-24.
Object id is 9606, canonical name is CompassAndStraightedgeConstructionOfSimilarTriangles.
Accessed 3839 times total.

Classification:
AMS MSC51M15 (Geometry :: Real and complex geometry :: Geometric constructions)
 51F99 (Geometry :: Metric geometry :: Miscellaneous)

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