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 Just find two! (Posted on 2005-04-27)
Given a triangle ABC, how can you find a point P on AC and a point Q on BC, such that AP=PQ=QB?

N.B. A construction method is sought, and only compass and straightedge are allowed.

 See The Solution Submitted by Federico Kereki Rating: 4.3333 (3 votes)

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 Solution | Comment 24 of 25 |
` `
`WOLOG let |AC| be less than or equal to |BC|.`
`Construct a circle with C as the center and |AC| as the radius.`
`Construct point D on line segment BC such that |DB| = |AC|.`
`Construct a line through D and parallel to AB intersecting the circle at point E (on the same side of AC as point B).`
`Point Q is the intersection of AE with BC.`
`Construct a line through Q and parallel to CE intersecting line AC at point P.`
`From similar triangles CAE and PAQ,`
`    |AE|   |AC|   |CE|    ---- = ---- = ----.    |AQ|   |AP|   |PQ|`
`From similar triangles AQB and EQD,`
`    |AE|   |DB|    ---- = ----.    |AQ|   |QB|`
`Therefore,`
`    |AC|   |CE|   |DB|    ---- = ---- = ----.    |AP|   |PQ|   |QB|`
`Since |AC| = |CE| = |DB|, we have`
`      |AP| = |PQ| = |QB|.`
`If you let E be the other intersection and perform`
`the rest of the construction, then you get a second`
`solution unless angle C is 60 degrees (AE and BC`
`would be parallel).`
` `

Edited on May 17, 2005, 12:03 am

Edited on May 17, 2005, 11:51 pm
 Posted by Bractals on 2005-05-14 00:12:51

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