compass and straightedge construction of similar triangles


Let a>0 and b>0. If line segmentsMathworldPlanetmath of lengths a and b are constructiblePlanetmathPlanetmath, 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 endpointsMathworldPlanetmath as C and D.

    .CD
  2. 2.

    Extend the line segment past both C and D

    ...CD
  3. 3.

    Erect the perpendicularPlanetmathPlanetmath to CD at C.

    ....CD
  4. 4.

    Use the compass to determine a point E on the erected perpendicular such that CE=1.

    ....CDE
  5. 5.

    Use the compass to determine a point F on CE such that CF=b.

    Note that the pictures indicate that b>1, but the exact same procedure works if 0<b1.

    ....CDEF
  6. 6.

    Connect the points D and E.

    ....CDEF
  7. 7.

    Copy the angle CDE at F to form similar trianglesMathworldPlanetmath. Label the intersectionMathworldPlanetmath of the constructed ray and 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.

    ....CDEFG

This construction is justified by the following:

  • Since the angle CFG was copied from the angle CED and the two trianglesMathworldPlanetmath share the angle DCE, then the two triangles CED and CFG are similarMathworldPlanetmathPlanetmath;

  • Since CEDCFG, we have that CECF=CDCG;

  • 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
Canonical name CompassAndStraightedgeConstructionOfSimilarTriangles
Date of creation 2013-03-22 17:16:02
Last modified on 2013-03-22 17:16:02
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 7
Author Wkbj79 (1863)
Entry type AlgorithmMathworldPlanetmath
Classification msc 51M15
Classification msc 51F99