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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?

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

 Subject Author Date Possible Solution Harry 2012-03-13 13:58:28 re(2): Initital exploration Jer 2012-03-10 00:34:58 re: Initital exploration Bractals 2012-03-09 15:09:04 Initital exploration Jer 2012-03-09 12:06:36

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