compass and straightedge construction of similar triangles
Let and . If line segments of lengths and are constructible, one can construct a line segment of length using compass and straightedge as follows:
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1.
Draw a line segment of length . Label its endpoints as and .
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2.
Extend the line segment past both and
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3.
Erect the perpendicular to at .
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4.
Use the compass to determine a point on the erected perpendicular such that .
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5.
Use the compass to determine a point on such that .
Note that the pictures indicate that , but the exact same procedure works if .
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6.
Connect the points and .
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7.
Copy the angle at to form similar triangles. Label the intersection of the constructed ray and as .
Note that, if , then will be between and , and will be between and . Also, if , then and .
This construction is justified by the following:
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Since , we have that ;
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Plugging in , , and yields that .
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 |