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Locus of Intersections (Posted on 2012-03-09) Difficulty: 3 of 5
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?

Prove it.

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


NOTATION:

  PQ denotes the vector from point P to point Q.

  b = |AC| and c = |AB|.

CLAIM:

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).

PROOF:

Let |x| = |BE| = |CD| with x < 0 if points D and E
are on the same side of the line BC as vertex A.
Then

  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
(1) gives

  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]

     and

  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.

QED

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionPossible SolutionHarry2012-03-13 13:58:28
re(2): Initital explorationJer2012-03-10 00:34:58
Some Thoughtsre: Initital explorationBractals2012-03-09 15:09:04
Initital explorationJer2012-03-09 12:06:36
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