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 20050514 00:12:51 