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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.

 See The Solution Submitted by Bractals Rating: 4.0000 (1 votes)

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 Possible Solution Comment 4 of 4 |
Taking the vertex, O, of the parallelogram ABOC as the origin, and denoting the position vectors of A, B... , by a, b ..., with magnitudes a, b..., it follows that:

e = b + (x/c)c   and   d = c + (x/b)b

where x is the (directed) distance of E from B and D from C.

If p and q denote CF as a proportion of CE and BF as a proportion of BD respectively, then
f = pe + (1 – p)c           =  p[b + (x/c)c] + (1 – p)c         (1)
and also:
f = qd + (1 – q)b           =  q[c + (x/b)b] + (1 - q)b

Equating coefficients of b and c in these two expressions for f:

p = qx/b  +  1  -  q         and       q = px/c  +  1  -  p

Solving simultaneously, these give           p = (x/b)/(x/c + x/b – x2/bc)

p = c/(b + c – x)            (x not 0)
Substituting in (1):

f    =  (c/(b + c – x))b + [(c/(b + c – x))(x/c – 1) + 1]c

= (1/(b + c – x))(cb + bc)

Since |cb| = |bc| = bc, it follows that f bisects the angle BOC and its magnitude depends on x. So, in relation to the original triangle ABC, F lies on a line parallel to the bisector of angle A and passes through the opposite vertex of the parallelogram ABOC.
When x = b + c (i.e. BE = AC + AB), there is a singularity giving F as a point at infinity. For x > b + c, f changes sign so that F describes the part of the locus on the other side of O.

 Posted by Harry on 2012-03-13 13:58:28

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