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|Locus of Intersections (Posted on 2012-03-09)
Let ABC be a triangle with points D and E lying on
lines AC and AB respectively such that D and E are
on the same side of line BC and |BE| = |CD| > 0. Let F be the intersection of rays BD and CE.
What is the locus of the intersections F?
Submitted by Bractals
Rating: 4.0000 (1 votes)
PQ denotes the vector from point P to point Q.
b = |AC| and c = |AB|.
Let G be the point on ray CA such that |CG| = c.
The locus of intersections is the line through G
and parallel to the bisector of ∠BAC (excluding
the point where the line intersects line BC).
Let |x| = |BE| = |CD| with x < 0 if points D and E
are on the same side of the line BC as vertex A.
AE = [(c+x)/c]AB AD = [(b+x)/b]AC
AF = (1-λ)AB + λAD
= (1-λ)AB + λ[(b+x)/b]AC (1)
AF = μAE + (1-μ)AC
= μ[(c+x)/c]AB + (1-μ)AC (2)
Solving (1) and (2) for λ and substituting into
AF = [(c+x)AB + (b+x)AC]/(b+c+x)
= [(b-c)/b]AC + [c(c+x)/(b+c+x)](AB/c + AC/b)
= AG + p[AB/c + AC/b]
Therefore, each intersection lies on the line
through G and parallel to the bisector of ∠BAC.
If P is a point on the line through G and parallel
to the bisector of ∠BAC, then
AP = AG + p[AB/c + AC/b]
x = [bp - c(c-p)]/(c-p)
will determine points D and E such that P is the
intersection of BD and CE.
Therefore, the claim holds.
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