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compass and straightedge construction of similar triangles
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(Algorithm)
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Let  and  . If line segments of lengths  and  are constructible, one can construct a line segment of length  using compass and straightedge as follows:
- Draw a line segment of length
 . Label its endpoints as  and  .
- Extend the line segment past both
 and 
- Erect the perpendicular to
 at  .
- Use the compass to determine a point
 on the erected perpendicular such that  .
- Use the compass to determine a point
 on
 such that  .
Note that the pictures indicate that  , but the exact same procedure works if  .
- Connect the points
 and  .
- 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:
- Since the angle
 was copied from the angle
 and the two triangles share the angle
 , then the two triangles
 and
 are similar;
- Since
 , we have that
 ;
- Plugging in
 ,  , and  yields that  .
If you are interested in seeing the rules for compass and straightedge constructions, click on the link provided.
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"compass and straightedge construction of similar triangles" is owned by Wkbj79.
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(view preamble)
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
There are 3 references to this entry.
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 1634 times total.
Classification:
| AMS MSC: | 51M15 (Geometry :: Real and complex geometry :: Geometric constructions) | | | 51F99 (Geometry :: Metric geometry :: Miscellaneous) |
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Pending Errata and Addenda
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