Transformation Geometry -- Math 331
January 30, 2004
Discussion
Theorem.
Let A, B, and C be non-collinear points, and let D,
E, and F, respectively, be given points on the lines
BC, CA, and AB, respectively, with corresponding
barycentric representations D = d'B + d''C, E = e'C + e''A,
and F = f'A + f''B and none of D, E, or F equal to
any of A, B, or C. Then the three
lines AD, BE, and CF meet in a common point if and only if
Proof.
Note that none of the numbers d', d'', e', e'', f', f'' can be zero
under the given hypotheses. If P = uA + vB + wC (with
u + v + w = 1) is a common point on AD, BE, and CF,
then by the principle of preservation
of proportionality in barycentric combinations (Problem 2, Assignment
due January 28) one has
d' = {v}/{v+w}, d'' = {w}/{v+w}, e' = {w}/{w+u}, e'' = {u}/{w+u}, f' = {u}/{u+v}, and f'' = {v}/{u+v} .
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Therefore, d'/d'' = v/w, e'/e'' = w/u, and f'/f'' = u/v,
and then clearly the product of these three values is 1.
Conversely, assume that (d'/d'')(e'/e'')(f'/f'') = 1. The
question now is whether there is a triple of numbers (u', v', w'),
none zero, such that the three pairs (v', w'), (w', u'), and
(u', v'), respectively, are parallel to (d', d''), (e',
e''), and (f', f''). For if that is the case, then with
u = {u'}/{u' + v' + w'}, v = {v'}/{u' + v' + w'}, and w = {w'}/{u' + v' + w'}
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the lines drawn from the point P = uA + vB + cW to the vertices
meet the sides of the triangle in the points D, E, F by the
principle of preservation
of proportionality in barycentric combinations. For the existence
of such u', v', w' let
u' = f', v' = f'', and w' = {d''f''}/{d'} .
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Then clearly (u', v') is parallel to (f', f'') and
(v', w') = (f''/d')(d', d'') is parallel to (d', d'').
Finally,
by formula (1) and, therefore, (w', u') is parallel
to (e', e'').
Exercises due Monday, February 2
For what values of c are the three points (c, -1),
(3, 2), and (-2, 1) barycentrically dependent? What is the
geometric significance of this issue?
Prove Ceva's Theorem: If P is a point in the interior
of the triangle determined by three non-collinear points A,
B, and C and if D, E, and F, respectively, are
the points where the lines from P to the points A, B, and
C, respectively, meet the lines BC, CA, and AB,
respectively, then one has the relation
{ |BD| }/{ |DC| }{ |CE| }/{ |EA| }{ |AF| }/{ |FB| } = 1
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among the lengths of the six line segments.
Prove: If A, B, and C are three non-collinear points
in a plane and l is a line in that plane meeting the lines BC,
CA, and AB, respectively in points D, E, and F,
respectively, having barycentric coordinate pairs (d', d''),
(e', e''), and (f', f''), respectively, with respect to
A, B, C, then one has the relation
AUTHOR
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COMMENT