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Similar Triangles (Posted on 2013-11-15) Difficulty: 3 of 5

  
Let ABCD be a simple quadrilateral.

Construct with straightedge and compass a point P
such that ΔPAB ~ ΔPCD and both triangles have the
same orientation.
  

  Submitted by Bractals    
Rating: 4.0000 (1 votes)
Solution: (Hide)

  
Let the geometric points be represented by complex
numbers.

                                                   A-P          C-P
   ΔPAB ~ ΔPCD (direct)  <==>  -------- = ---------
                                                   B-P          D-P

Solving for P gives

                 BC - AD
   P = ---------------------
            (B+C) - (A+D)

The denominator equals zero if and only if the
midpoints of sides AD and BC coincide. Since
ABCD is a simple quadrilateral, point P exists.

WOLOG let A=0 and B=1, then

   P = C/(C-D+1).

Let F = C-D+1, then P = C/F.

Construct the point F using vectors,

   AF = AC - AD + AB.

Construct point P by constructing ΔABP such that
ΔABP ~ ΔAFC (Direct).

QED
  

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(6): Argand - grand!Harry2013-12-04 14:09:41
re(5): Argand - grand!Bractals2013-12-04 05:14:19
re(4): Argand - grand!Harry2013-12-03 20:28:08
re(3): Argand - grand!Bractals2013-12-03 17:39:31
re(2): Argand - grand!Harry2013-12-03 12:47:12
re: Argand - grand!Bractals2013-11-28 22:36:03
Argand - grand!Harry2013-11-27 14:27:13
re(2): SolutionHarry2013-11-24 18:31:59
re: SolutionBractals2013-11-24 14:59:58
SolutionSolutionHarry2013-11-23 20:44:38
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