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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?
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)
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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