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:
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1.
Draw a line segment of length a. Label its endpoints as C and D.
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2.
Extend the line segment past both C and D
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3.
Erect the perpendicular to ↔CD at C.
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4.
Use the compass to determine a point E on the erected perpendicular such that CE=1.
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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<b≤1.
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6.
Connect the points D and E.
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7.
Copy the angle ∠CDE at F to form similar triangles. Label the intersection 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.
This construction is justified by the following:
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Since △CED∼△CFG, we have that CECF=CDCG;
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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 |
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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 | Algorithm |
Classification | msc 51M15 |
Classification | msc 51F99 |