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.
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 |
Solution
