Home > Shapes > Geometry
Similar Triangles (Posted on 2013-11-15) |
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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.
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Submitted by Bractals
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Rating: 4.0000 (1 votes)
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Solution:
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(Hide)
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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
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